Ch 2  ·  Q–
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Chapter 2 Miscellaneous Exercise Solutions

Inverse Trigonometric Functions

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 2 Miscellaneous Exercise

Class 12 Mathematics Miscellaneous Exercise NCERT Solutions Inverse Trigonometric Functions Class 12 Mathematics Chapter 2 CBSE Board Exam JEE Main CUET Principal Values Inverse Sine Inverse Cosine Inverse Tangent Inverse Secant Inverse Cosecant Inverse Cotangent
14 Questions
30–45 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
Find the value of \(\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)\)
📘 Concept & Theory
Concept/Theory
›

The function \(\cos^{-1}x\), also called the inverse cosine function, gives the principal value of the angle whose cosine is \(x\).

The principal value range of \(\cos^{-1}x\) is

\[ 0\leq\cos^{-1}x\leq\pi. \]

Therefore, while evaluating an expression of the form \(\cos^{-1}(\cos\theta)\), we cannot always directly write \(\cos^{-1}(\cos\theta)=\theta\). This is valid only when \(\theta\) lies in the principal value range \([0,\pi]\).

If the given angle lies outside this interval, it must first be reduced to an equivalent angle whose value lies in the principal range of \(\cos^{-1}x\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite \(\dfrac{13\pi}{6}\) as a full revolution plus a standard angle.

  2. Use the periodicity of the cosine function.

  3. Evaluate the resulting standard cosine value.

  4. Apply the principal value range of \(\cos^{-1}x\).

  5. Obtain the required principal value.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. We have
    \[\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)\]
  2. First, express the angle \(\dfrac{13\pi}{6}\) as
    \[\begin{aligned}\dfrac{13\pi}{6}&=\dfrac{12\pi}{6}+\dfrac{\pi}{6}\\ &=2\pi+\dfrac{\pi}{6}\end{aligned}\]
  3. Therefore, \(\cos\dfrac{13\pi}{6}=\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}\).
  4. Since cosine is periodic with period \(2\pi\),
    \[\cos(\theta+2\pi)=\cos\theta\]
  5. Hence,
    \[\cos\left(2\pi+\dfrac{\pi}{6}\right)=\cos\dfrac{\pi}{6}\]
  6. Using the standard trigonometric value,
    \[\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\]
  7. Thus, the given expression becomes
    \[\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\]
  8. We know that
    \[\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\]
  9. Since,
    \[\dfrac{\pi}{6}\in[0,\pi],\]
    it belongs to the principal value range of \(\cos^{-1}x\). Therefore
    \[\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\pi}{6}\]
  10. Hence,
    \[\boxed{\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)=\dfrac{\pi}{6}}\]
🎯 Exam Significance
Exam Significance
›

This problem tests a fundamental concept of inverse trigonometric functions: the distinction between a trigonometric angle and the principal value of its inverse trigonometric function.

In board examinations, questions involving \(\sin^{-1}(\sin\theta)\), \(\cos^{-1}(\cos\theta)\), and \(\tan^{-1}(\tan\theta)\) frequently test whether the student knows the corresponding principal value ranges. A direct cancellation of the trigonometric function and its inverse without checking the principal range can lead to an incorrect answer.

Significance for Competitive Entrance Examinations

For competitive examinations, this concept is particularly important because inverse trigonometric expressions are often combined with angles outside their principal ranges. The ability to reduce an angle using periodicity and then select the correct principal value allows such questions to be solved quickly and reliably.

The key point is that

\[ \cos^{-1}(\cos\theta)=\theta \]

only when

\[ 0\leq\theta\leq\pi. \]

For angles outside this interval, the angle must first be transformed to the corresponding angle in the principal value range.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  1 point
  1. 🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  6 points

    1. The principal value range of \(\cos^{-1}x\) is \([0,\pi]\).

    2. The cosine function has period \(2\pi\).

    3. \(\dfrac{13\pi}{6}=2\pi+\dfrac{\pi}{6}\).

    4. Therefore, \(\cos\dfrac{13\pi}{6}=\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\).

    5. The principal value of \(\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\) is \(\dfrac{\pi}{6}\).

    6. Always check the principal value range before simplifying an inverse trigonometric expression.
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1 / 14  ·  7%
Q2 →
Q2
NUMERIC3 marks
Find the value of $\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right)$
📘 Concept & Theory
Concept/Theory
›

The inverse tangent function \(\tan^{-1}x\) gives the principal value of the angle whose tangent is \(x\).

The principal value range of \(\tan^{-1}x\) is

\[ -\dfrac{\pi}{2}<\tan^{-1}x<\dfrac{\pi}{2}. \]

Therefore, while evaluating an expression of the form \(\tan^{-1}(\tan\theta)\), we cannot always directly write \(\tan^{-1}(\tan\theta)=\theta\).

This direct cancellation is valid only when

\[ -\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}. \]

If \(\theta\) lies outside this principal range, we must find a coterminal angle that lies within the principal range of \(\tan^{-1}x\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Express \(\dfrac{7\pi}{6}\) as an angle plus \(\pi\).

  2. Use the periodicity of the tangent function.

  3. Reduce the angle to an equivalent angle within the principal range of \(\tan^{-1}x\).

  4. Apply the inverse tangent function.

  5. Write the final principal value.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  12 steps
  1. We have
    \[\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right)\]
  2. First, express \(\dfrac{7\pi}{6}\) as
    \[ \dfrac{7\pi}{6}=\pi+\dfrac{\pi}{6}\]
  3. Therefore,
    \[\tan\dfrac{7\pi}{6}=\tan\left(\pi+\dfrac{\pi}{6}\right)\]
  4. The tangent function has period \(\pi\), so
    \[\tan(\theta+\pi)=\tan\theta\]
  5. Hence,
    \[\tan\left(\pi+\dfrac{\pi}{6}\right)=\tan\dfrac{\pi}{6}\]
  6. Using the standard trigonometric value,
    \[\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}\]
  7. Thus,
    \[\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right) = \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\]
  8. We know that
    \[\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}\]
  9. Also,
    \[-\dfrac{\pi}{2} < \dfrac{\pi}{6} < \dfrac{\pi}{2}\]
  10. Therefore, \(\dfrac{\pi}{6}\) lies within the principal value range of \(\tan^{-1}x\)
  11. Hence,
    \[\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\dfrac{\pi}{6}\]
  12. Therefore,
    \[\boxed{\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right)=\dfrac{\pi}{6}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the important concept of principal values of inverse trigonometric functions. In board examinations, students are expected to distinguish between the periodicity of the original trigonometric function and the restricted range of its inverse function.

A common error is to cancel \(\tan^{-1}\) and \(\tan\) directly and write \(\dfrac{7\pi}{6}\). This is incorrect because \(\dfrac{7\pi}{6}\) does not lie in the principal value range

\[ -\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}. \]

Significance for Competitive Entrance Examinations

This type of problem is useful for developing speed and accuracy in competitive examinations such as JEE and other entrance tests. Questions involving inverse trigonometric functions often deliberately use angles outside the principal range to test whether the candidate understands periodicity and principal values.

The key relationship to remember is

\[ \tan(\theta+\pi)=\tan\theta, \]

together with the principal value range

\[ -\dfrac{\pi}{2} < \tan^{-1}x < \dfrac{\pi}{2}. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The period of the tangent function is \(\pi\).

  2. \(\dfrac{7\pi}{6}=\pi+\dfrac{\pi}{6}\).

  3. Therefore, \(\tan\dfrac{7\pi}{6}=\tan\dfrac{\pi}{6}\).

  4. \(\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}\).

  5. The principal value range of \(\tan^{-1}x\) is \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\).

  6. Since \(\dfrac{\pi}{6}\) lies in this range, \(\tan^{-1}\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}\).

  7. Do not directly cancel \(\tan^{-1}\) and \(\tan\) unless the given angle lies within the principal value range of \(\tan^{-1}x\).
← Q1
2 / 14  ·  14%
Q3 →
Q3
NUMERIC3 marks
Prove that $2\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}$
📘 Concept & Theory
Concept/Theory
›

This problem uses the relationship between inverse trigonometric functions, half-angle identities, and the double-angle formula for tangent.

We begin by setting the inverse sine expression equal to an angle. Since

\[ \sin^{-1}\dfrac{3}{5} \]
is a principal value, we have

\[ 0\leq\sin^{-1}\dfrac{3}{5}\leq\dfrac{\pi}{2}. \]

Consequently, if

\[ \theta=2\sin^{-1}\dfrac{3}{5}, \]
then

\[ 0\leq\dfrac{\theta}{2}\leq\dfrac{\pi}{2}. \]

Therefore, \(\cos\dfrac{\theta}{2}\) is positive, which is important when evaluating it using the identity

\[ \cos^2\dfrac{\theta}{2} = 1-\sin^2\dfrac{\theta}{2}. \]

After finding

\[ \tan\dfrac{\theta}{2}, \]
we use the double-angle identity

\[ \tan\theta = \dfrac{2\tan(\theta/2)} {1-\tan^2(\theta/2)}. \]

Finally, we use the principal value range of \(\tan^{-1}x\):

\[ -\dfrac{\pi}{2}<\tan^{-1}x<\dfrac{\pi}{2}. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(2\sin^{-1}\dfrac{3}{5}=\theta\).

  2. Obtain \(\sin\dfrac{\theta}{2}=\dfrac{3}{5}\).

  3. Find \(\cos\dfrac{\theta}{2}\) using the Pythagorean identity.

  4. Calculate \(\tan\dfrac{\theta}{2}\).

  5. Apply the double-angle formula for \(\tan\theta\).

  6. Show that \(\tan\theta=\dfrac{24}{7}\).

  7. Use the principal value range to conclude that \(\theta=\tan^{-1}\dfrac{24}{7}\).

  8. Substitute the value of \(\theta\) to establish the required result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  20 steps
  1. Let
    \[\theta=2\sin^{-1}\dfrac{3}{5}\]
  2. Dividing both sides by \(2\), we get
    \[\dfrac{\theta}{2}=\sin^{-1}\dfrac{3}{5}\]
  3. Taking sine on both sides,
    \[\sin\dfrac{\theta}{2}=\dfrac{3}{5}\]
  4. Since
    \[\dfrac{\theta}{2}=\sin^{-1}\dfrac{3}{5},\]
    and the principal value range of \(\sin^{-1}x\) is
    \[-\dfrac{\pi}{2}\leq\sin^{-1}x\leq\dfrac{\pi}{2}\]
    while \(\dfrac{3}{5}>0\), we have
    \[0 < \dfrac{\theta}{2} < \dfrac{\pi}{2}\]
  5. Hence,
    \[\cos\dfrac{\theta}{2}>0\]
  6. Using
    \[ \sin^2\dfrac{\theta}{2}+\cos^2\dfrac{\theta}{2}=1\]
  7. we get
    \[ \cos^2\dfrac{\theta}{2}=1-\sin^2\dfrac{\theta}{2}\]
  8. Substituting
    \[\sin\dfrac{\theta}{2}=\dfrac{3}{5}\]
  9. we obtain
    \[\begin{aligned} \cos^2\dfrac{\theta}{2}&=1-\left(\dfrac{3}{5}\right)^2\\ &=1-\dfrac{9}{25}\\ &=\dfrac{25-9}{25}\\ &=\dfrac{16}{25} \end{aligned}\]

    Since \(\cos\dfrac{\theta}{2}>0\),

    \[\begin{aligned} \cos\dfrac{\theta}{2}&=\sqrt{\dfrac{16}{25}}\\ &=\dfrac{4}{5}\end{aligned}\]
  10. Therefore,
    \[\tan\dfrac{\theta}{2}=\dfrac{\sin(\theta/2)}{\cos(\theta/2)}\]
  11. Substituting the values obtained above,
    \[\begin{aligned}\tan\dfrac{\theta}{2}&=\dfrac{3/5}{4/5}\\&=\dfrac{3}{5}\times\dfrac{5}{4}\\&=\dfrac{3}{4}\end{aligned}\]
  12. using the double-angle formula
    \[\tan\theta=\dfrac{2\tan(\theta/2)}{1-\tan^2(\theta/2)}\]
  13. we get
    \[\begin{aligned}\tan\theta&=\dfrac{2\left(\dfrac{3}{4}\right)}{1-\left(\dfrac{3}{4}\right)^2}\\ &=\dfrac{\dfrac{3}{2}}{1-\dfrac{9}{16}}\\ & =\dfrac{\dfrac{3}{2}}{\dfrac{16-9}{16}}\\ &=\dfrac{\dfrac{3}{2}}{\dfrac{7}{16}}\\ &=\dfrac{3}{2}\times\dfrac{16}{7}\\ &=\dfrac{48}{14}\\ &=\dfrac{24}{7}\end{aligned}\]
  14. Thus,
    \[\tan\theta=\dfrac{24}{7}\]
  15. Taking \(\tan^{-1}\) on both sides gives
    \[\tan^{-1}(\tan\theta)=\tan^{-1}\dfrac{24}{7}\]
  16. From
    \[0 < \dfrac{\theta}{2} < \dfrac{\pi}{2}\]
    it follows that
    \[0 < \theta < \pi\]
  17. In fact,
    \[ \tan\theta=\dfrac{24}{7}>0. \]
    Since \(\theta\) lies between \(0\) and \(\pi\), a positive tangent implies that \(\theta\) lies in the first quadrant. Hence,
    \[0 < \theta < \dfrac{\pi}{2}\]
  18. Therefore, \(\theta\) lies within the principal value range of \(\tan^{-1}x\), which is
    \[-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\]
  19. Consequently,
    \[\theta=\tan^{-1}\dfrac{24}{7}\]
  20. But from our initial definition,
    \[\theta=2\sin^{-1}\dfrac{3}{5}\]
  21. Therefore,
    \[\boxed{2\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}}\]
  22. Hence Proved
🎯 Exam Significance
Exam Significance
›

This problem combines several important concepts from the chapter: principal values of inverse trigonometric functions, the Pythagorean identity, half-angle reasoning, and the double-angle formula for tangent. Such questions are important for testing whether the student can connect different trigonometric identities rather than applying a single formula.

The principal value condition is particularly important. After obtaining \(\tan\theta=\dfrac{24}{7}\), it is necessary to justify why

\[ \theta=\tan^{-1}\dfrac{24}{7} \]
rather than simply applying \(\tan^{-1}\) without considering the range. This makes the proof mathematically complete and is important in a board-examination solution.

Significance for Competitive Entrance Examinations

For competitive entrance examinations, this problem develops the ability to transform an inverse trigonometric expression into an ordinary trigonometric ratio. The values \(3\), \(4\), and \(5\) form a Pythagorean triple, making the half-angle calculation particularly efficient.

A useful pattern to recognize is

\[ \sin\alpha=\dfrac{3}{5} \quad\Longrightarrow\quad \cos\alpha=\dfrac{4}{5} \quad\Longrightarrow\quad \tan\alpha=\dfrac{3}{4}. \]

The double-angle formula then immediately gives

\[ \tan 2\alpha = \dfrac{2(3/4)} {1-(3/4)^2} = \dfrac{24}{7}. \]

Recognising this structure can substantially reduce the time required to solve similar problems in entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. When \( \theta=2\sin^{-1}\dfrac{3}{5} \), we have \( \sin\dfrac{\theta}{2}=\dfrac{3}{5} \).

  2. Because \(\dfrac{\theta}{2}\) is a principal inverse-sine value, \(\cos\dfrac{\theta}{2}\) is positive.

  3. \(\cos\dfrac{\theta}{2}=\dfrac{4}{5}\).

  4. \(\tan\dfrac{\theta}{2}=\dfrac{3}{4}\).

  5. The correct double-angle identity is \(\tan\theta=\dfrac{2\tan(\theta/2)}{1-\tan^2(\theta/2)}\).

  6. Substitution gives \(\tan\theta=\dfrac{24}{7}\).

  7. The range of \(\tan^{-1}x\) is \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\).

  8. Since \(\theta\) is in the first quadrant, \(\theta=\tan^{-1}\dfrac{24}{7}\).

  9. Therefore, \(2\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}\).
← Q2
3 / 14  ·  21%
Q4 →
Q4
NUMERIC3 marks
Prove that \[\sin^{-1}\dfrac{8}{17}+\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{77}{36}\]
📘 Concept & Theory
Concept/Theory
›

This problem uses the conversion of inverse sine expressions into inverse tangent expressions, followed by the tangent addition formula.

If

\[ \theta=\sin^{-1}x, \]
then
\[ \sin\theta=x. \]
Since the principal value range of \(\sin^{-1}x\) is
\[ -\dfrac{\pi}{2}\leq\theta\leq\dfrac{\pi}{2}, \]
the sign of \(\cos\theta\) must be considered before finding \(\tan\theta\).

Here both inverse sine values are positive. Therefore, both corresponding angles lie in the first quadrant, so their cosine values are positive.

After finding the tangent of each angle, we use the tangent addition formula

\[ \tan(A+B) = \dfrac{\tan A+\tan B} {1-\tan A\tan B}. \]

Finally, the principal value condition must be checked. Since both angles are positive and their sum is less than \(\dfrac{\pi}{2}\), the sum itself lies in the principal value range of \(\tan^{-1}x\). Hence, we may conclude that the sum is equal to the corresponding inverse tangent value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(\theta_1=\sin^{-1}\dfrac{8}{17}\).

  2. Find \(\cos\theta_1\) and then \(\tan\theta_1\).

  3. Let \(\theta_2=\sin^{-1}\dfrac{3}{5}\).

  4. Find \(\cos\theta_2\) and then \(\tan\theta_2\).

  5. Apply the tangent addition formula to \(\theta_1+\theta_2\).

  6. Simplify the resulting tangent to \(\dfrac{77}{36}\).

  7. Verify that \(\theta_1+\theta_2\) lies in the principal range of \(\tan^{-1}x\).

  8. Take \(\tan^{-1}\) to obtain the required result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  38 steps
  1. Let
    \[\theta_1=\sin^{-1}\dfrac{8}{17}\]
  2. Therefore,
    \[\sin\theta_1=\dfrac{8}{17}\]
  3. Since
    \[ 0 < \dfrac{8}{17} < 1, \]
    we have
    \[ 0 < \theta_1 < \dfrac{\pi}{2}. \]
    Hence, \(\cos\theta_1>0\)
  4. \[\sin^2\theta_1+\cos^2\theta_1=1\]
  5. we get
    \[ \cos^2\theta_1 = 1-\sin^2\theta_1. \]
  6. Substituting \(\sin\theta_1=\dfrac{8}{17}\),
  7. \[\begin{aligned} \cos^2\theta_1&=1-\left(\dfrac{8}{17}\right)^2\\ &=1-\dfrac{64}{289}\\ &=\dfrac{289-64}{289}\\ &=\dfrac{225}{289} \end{aligned} \]
  8. Since \(\cos\theta_1>0\),
    \[\begin{aligned} \cos\theta_1&=\sqrt{\dfrac{225}{289}}\\ &=\dfrac{15}{17} \end{aligned}\]
  9. Therefore,
    \[\begin{aligned} \tan\theta_1&=\dfrac{\sin\theta_1}{\cos\theta_1}\\ &=\dfrac{8/17}{15/17}\\ &=\dfrac{8}{15} \end{aligned} \]
  10. Thus,
    \[\tan\theta_1=\dfrac{8}{15}\]
  11. Since
    \[0 < \theta_1 < \dfrac{\pi}{2}\]
  12. we may write
    \[\theta_1=\tan^{-1}\dfrac{8}{15}\]
  13. Hence,
    \[\sin^{-1}\dfrac{8}{17}=\tan^{-1}\dfrac{8}{15}\]
  14. Now let
    \[\theta_2=\sin^{-1}\dfrac{3}{5}\]
  15. Therefore,
    \[\sin\theta_2=\dfrac{3}{5}\]
  16. Since
    \[0 < \dfrac{3}{5} < 1,\]
    we have
    \[0 < \theta_2 < \dfrac{\pi}{2}\]
    Hence, \(\cos\theta_2>0\)
  17. Using
    \[\sin^2\theta_2+\cos^2\theta_2=1\]
  18. we obtain
    \[\cos^2\theta_2=1-\sin^2\theta_2\]
  19. Substituting \(\sin\theta_2=\dfrac{3}{5}\),
  20. \[\begin{aligned} \cos^2\theta_2&=1-\left(\dfrac{3}{5}\right)^2\\ &=1-\dfrac{9}{25}\\ &=\dfrac{16}{25} \end{aligned} \]
  21. Since \(\cos\theta_2>0\),
    \[\begin{aligned}\cos\theta_2&=\sqrt{\dfrac{16}{25}} &=\dfrac{4}{5}\end{aligned}\]
  22. Therefore,
    \[\begin{aligned} \tan\theta_2&=\dfrac{\sin\theta_2}{\cos\theta_2}\\ &=\dfrac{3/5}{4/5}\\ &=\dfrac{3}{4} \end{aligned} \]
  23. Thus,
    \[\tan\theta_2=\dfrac{3}{4}\]
  24. Since
    \[0 < \theta_2 < \dfrac{\pi}{2}\]
  25. we also have
    \[\theta_2=\tan^{-1}\dfrac{3}{4}\]
  26. Now consider the sum
    \[\theta_1+\theta_2\]
  27. Using the tangent addition formula,
    \[\tan(\theta_1+\theta_2)=\dfrac{\tan\theta_1+\tan\theta_2}{1-\tan\theta_1\tan\theta_2}\]
  28. Substituting
    \[\tan\theta_1=\dfrac{8}{15}\quad\text{and}\]
    \[\tan\theta_2=\dfrac{3}{4}\]
  29. we get
    \[\begin{aligned}\tan(\theta_1+\theta_2) &= \dfrac{\dfrac{8}{15}+\dfrac{3}{4}} {1-\left(\dfrac{8}{15}\right)\left(\dfrac{3}{4}\right)} &\end{aligned}\]
  30. Taking the LCM of \(15\) and \(4\) in the numerator,
    \[\begin{aligned} \dfrac{8}{15}+\dfrac{3}{4} &= \dfrac{32}{60}+\dfrac{45}{60}\\ &=\dfrac{77}{60} \end{aligned}\]
  31. Now simplify the denominator:
    \[\begin{aligned} 1-\left(\dfrac{8}{15}\right)\left(\dfrac{3}{4}\right) &=1-\dfrac{24}{60}\\ &=1-\dfrac{2}{5}\\ &=\dfrac{5-2}{5}\\ &=\dfrac{3}{5} \end{aligned} \]
  32. Therefore,
    \[\tan(\theta_1+\theta_2)=\dfrac{77/60}{3/5}\]
  33. Dividing by \(\dfrac{3}{5}\),
    \[\begin{aligned} \tan(\theta_1+\theta_2)&=\dfrac{77}{60}\times\dfrac{5}{3}\\ &=\dfrac{77}{36}\end{aligned}\]
  34. Hence,
    \[\tan(\theta_1+\theta_2)=\dfrac{77}{36}\]
  35. Now, since
    \[0 < \theta_1 < \dfrac{\pi}{2}\quad\text{and}\]
    \[0 < \theta_2 < \dfrac{\pi}{2}\]
  36. we know that
    \[0 < \theta_1+\theta_2 < \pi\]
  37. Also,
    \[ \tan(\theta_1+\theta_2)=\dfrac{77}{36}>0. \]
    Since the sum lies between \(0\) and \(\pi\) and its tangent is positive, the sum must lie in the first quadrant. Therefore,
    \[0 < \theta_1+\theta_2 <\ dfrac{\pi}{2}\]
  38. Thus, \(\theta_1+\theta_2\) lies in the principal value range of \(\tan^{-1}x\), namely
    \[-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\]
  39. Hence,
    \[\theta_1+\theta_2=\tan^{-1}\dfrac{77}{36}\]
  40. Using
    \[\theta_1=\sin^{-1}\dfrac{8}{17}\quad\text{and}\]
    \[\theta_2=\sin^{-1}\dfrac{3}{5},\]
  41. we obtain
    \[\boxed{\sin^{-1}\dfrac{8}{17}+\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{77}{36}}\]
  42. Hence, Proved
🎯 Exam Significance
Exam Significance
›

This problem is important for board examinations because it combines inverse trigonometric functions with the tangent addition formula. It tests whether the student can correctly determine the signs of the cosine values, calculate tangent ratios, and handle principal values.

The principal-value justification is an important part of a complete proof. Simply obtaining

\[ \tan(\theta_1+\theta_2)=\dfrac{77}{36} \]
is not by itself sufficient to conclude the required inverse-tangent equality. The range of the angle must also be considered.

Significance for Competitive Entrance Examinations

This problem provides an efficient pattern for competitive examinations. The pairs

\[ 8,15,17 \]
and
\[ 3,4,5 \]
are Pythagorean triples. Therefore, once the inverse sine values are converted into angles, their tangent values can be obtained rapidly:

\[ \sin\theta_1=\dfrac{8}{17} \quad\Longrightarrow\quad \tan\theta_1=\dfrac{8}{15}, \]
\[ \sin\theta_2=\dfrac{3}{5} \quad\Longrightarrow\quad \tan\theta_2=\dfrac{3}{4}. \]

The tangent addition formula then reduces the problem to straightforward rational-number simplification. Recognising Pythagorean triples can significantly improve speed in JEE and other competitive entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(\sin^{-1}\dfrac{8}{17}\) corresponds to a first-quadrant angle with \(\cos\theta_1=\dfrac{15}{17}\).

  2. Therefore, \(\tan\theta_1=\dfrac{8}{15}\).

  3. \(\sin^{-1}\dfrac{3}{5}\) corresponds to a first-quadrant angle with \(\cos\theta_2=\dfrac{4}{5}\).

  4. Therefore, \(\tan\theta_2=\dfrac{3}{4}\).

  5. The correct tangent addition formula is \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\).

  6. \(\tan(\theta_1+\theta_2)=\dfrac{77}{36}\).

  7. The sign of the tangent and the range of the sum establish that \(\theta_1+\theta_2\) lies in the principal range of \(\tan^{-1}\).

  8. The triples \(8,15,17\) and \(3,4,5\) are useful for quickly identifying the required cosine and tangent values.
← Q3
4 / 14  ·  29%
Q5 →
Q5
NUMERIC3 marks
Prove that \(\cos^{-1}\dfrac{4}{5}+\cos^{-1}\dfrac{12}{13}=\cos^{-1}\dfrac{33}{65}\)
📘 Concept & Theory
Concept/Theory
›

This problem uses the cosine addition formula together with the principal value properties of the inverse cosine function.

If

\[ A=\cos^{-1}x, \]
then
\[ \cos A=x \]
and the principal value of \(A\) lies in
\[ 0\leq A\leq\pi. \]

Since both \(\dfrac{4}{5}\) and \(\dfrac{12}{13}\) are positive, the angles \(A\) and \(B\) lie in the first quadrant. Therefore, their sine values are positive.

We then use the cosine addition formula

\[ \cos(A+B)=\cos A\cos B-\sin A\sin B. \]

After finding \(\cos(A+B)\), we must verify that \(A+B\) is within the principal value range of \(\cos^{-1}\), namely

\[ 0\leq A+B\leq\pi. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(A=\cos^{-1}\dfrac{4}{5}\).

  2. Find \(\sin A\) using \(\sin^2A+\cos^2A=1\).

  3. Let \(B=\cos^{-1}\dfrac{12}{13}\).

  4. Find \(\sin B\) using \(\sin^2B+\cos^2B=1\).

  5. Apply the cosine addition formula to \(A+B\).

  6. Simplify \(\cos(A+B)\) to \(\dfrac{33}{65}\).

  7. Check the range of \(A+B\) so that the inverse cosine can be applied correctly.

  8. Conclude the required identity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  22 steps
  1. Let
    \[A=\cos^{-1}\dfrac{4}{5}\]
  2. Therefore,
    \[\cos A=\dfrac{4}{5}\]
  3. Since
    \[0<\dfrac{4}{5}<1\]
    , we have
    \[0 Hence, \(\sin A>0\)
  4. Using the identity
    \[\sin^2A+\cos^2A=1\]
  5. we get
    \[\sin^2A=1-\cos^2A\]
  6. Substituting \(\cos A=\dfrac{4}{5}\),
    \[\begin{aligned} \sin^2A &=1-\left(\dfrac{4}{5}\right)^2\\ &=1-\dfrac{16}{25}\\ &=\dfrac{25-16}{25}\\ &=\dfrac{9}{25} \end{aligned} \]
  7. Since \(A\) is in the first quadrant,
    \[\begin{aligned}\sin A&=\sqrt{\dfrac{9}{25}}\\ &=\dfrac{3}{5}\end{aligned}\]
  8. Thus,
    \[\boxed{\sin A=\dfrac{3}{5}}\]
  9. Now let
    \[B=\cos^{-1}\dfrac{12}{13}\]
  10. Therefore,
    \[\cos B=\dfrac{12}{13}\]
  11. Since
    \[0 < \dfrac{12}{13} < 1\]
    , we have
    \[0 < B < \dfrac{\pi}{2}\]
    Hence, \(\sin B>0\)
  12. Using
    \[\sin^2B+\cos^2B=1\]
  13. we obtain
    \[\sin^2B=1-\cos^2B\]
  14. Substituting \(\cos B=\dfrac{12}{13}\),
    \[\begin{aligned} \sin^2B &= 1-\left(\dfrac{12}{13}\right)^2\\ &=1-\dfrac{144}{169}\\ &=\dfrac{169-144}{169}\\ &=\dfrac{25}{169} \end{aligned} \]
  15. Since \(B\) is in the first quadrant,
    \[\sin B=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}\]
  16. Thus,
    \[\boxed{\sin B=\dfrac{5}{13}}\]
  17. Now apply the cosine addition formula:
    \[\cos(A+B)=\cos A\cos B-\sin A\sin B\]
  18. Substituting the values of \(\cos A\), \(\cos B\), \(\sin A\), and \(\sin B\),
    \[\cos(A+B)=\left(\dfrac{4}{5}\right)\left(\dfrac{12}{13}\right)-\left(\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right)\]
  19. Multiplying the fractions,
    \[\begin{aligned} \cos(A+B) & =\dfrac{48}{65}-\dfrac{15}{65}\\ &=\dfrac{48-15}{65}\\ &=\dfrac{33}{65}\end{aligned} \]
  20. Hence,
    \[\cos(A+B)=\dfrac{33}{65}\]
  21. Since
    \[0 < A < \dfrac{\pi}{2}\]
    and
    \[0 < B < \dfrac{\pi}{2}\]
    we have
    \[0 < A+B < \pi\]
  22. Moreover,
    \[\cos(A+B)=\dfrac{33}{65}>0\]
    Since \(A+B\) lies between \(0\) and \(\pi\) and its cosine is positive, \(A+B\) lies in the first quadrant. Therefore,
    \[0 < A+B < \dfrac{\pi}{2}\]
  23. Thus, \(A+B\) lies within the principal value range of \(\cos^{-1}x\), which is
    \[0\leq\cos^{-1}x\leq\pi\]
  24. Therefore,
    \[A+B=\cos^{-1}\dfrac{33}{65}\]
  25. Substituting the values of \(A\) and \(B\),
    \[\boxed{\cos^{-1}\dfrac{4}{5}+\cos^{-1}\dfrac{12}{13}=\cos^{-1}\dfrac{33}{65}}\]
  26. Hence, Proved
🎯 Exam Significance
Exam Significance
›

This is an important proof-based question because it combines the principal values of inverse trigonometric functions with the cosine addition formula. A complete board-examination solution should not only calculate \(\cos(A+B)\), but should also establish that the resulting angle is in the appropriate principal range.

The Pythagorean triples

\[ 3,4,5 \]
and
\[ 5,12,13 \]
make the calculation of the sine values straightforward. Recognising these triples can make the solution both systematic and efficient.

Significance for Competitive Entrance Examinations

This problem is useful for competitive examinations because it demonstrates a fast conversion technique: when the cosine of an angle is given through a familiar Pythagorean triple, the corresponding sine can be identified immediately.

Here,

\[ \cos A=\dfrac{4}{5} \quad\Longrightarrow\quad \sin A=\dfrac{3}{5}, \]
\[ \cos B=\dfrac{12}{13} \quad\Longrightarrow\quad \sin B=\dfrac{5}{13}. \]

The cosine addition formula then gives the result with minimal computation. Such recognition is valuable in time-constrained entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The principal value range of \(\cos^{-1}x\) is \([0,\pi]\).

  2. If \(A=\cos^{-1}\dfrac{4}{5}\), then \(A\) lies in the first quadrant and \(\sin A=\dfrac{3}{5}\).

  3. If \(B=\cos^{-1}\dfrac{12}{13}\), then \(B\) lies in the first quadrant and \(\sin B=\dfrac{5}{13}\).

  4. The cosine addition formula is \(\cos(A+B)=\cos A\cos B-\sin A\sin B\).

  5. \(\cos(A+B)=\dfrac{33}{65}\).

  6. Because \(A+B\) is in the principal range of \(\cos^{-1}\), we can conclude \(A+B=\cos^{-1}\dfrac{33}{65}\).

  7. The Pythagorean triples \(3,4,5\) and \(5,12,13\) provide the required sine and cosine values efficiently.
← Q4
5 / 14  ·  36%
Q6 →
Q6
NUMERIC3 marks
Prove that \(\cos^{-1}\dfrac{12}{13}+\sin^{-1}\dfrac{3}{5}=\sin^{-1}\dfrac{56}{65}\)
📘 Concept & Theory
Concept/Theory
›

This problem uses the sine addition formula together with the principal value properties of inverse trigonometric functions.

We use

\[ \sin(A+B)=\sin A\cos B+\cos A\sin B. \]

The given inverse trigonometric functions allow us to determine the corresponding sine and cosine values. Since both given ratios are positive, the associated principal angles lie in the first quadrant, so the positive square root must be selected when finding the remaining trigonometric ratio.

After finding \(\sin(A+B)\), we must also verify that \(A+B\) lies in the principal value range of \(\sin^{-1}x\):

\[ -\dfrac{\pi}{2}\leq A+B\leq\dfrac{\pi}{2}. \]

In this problem, both angles are positive and their sum is less than \(\dfrac{\pi}{2}\), so the inverse sine can be applied directly to obtain the required result.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(A=\cos^{-1}\dfrac{12}{13}\).

  2. Find \(\sin A\) using the Pythagorean identity.

  3. Let \(B=\sin^{-1}\dfrac{3}{5}\).

  4. Find \(\cos B\) using the Pythagorean identity.

  5. Apply the sine addition formula to \(A+B\).

  6. Substitute the four required trigonometric values.

  7. Simplify \(\sin(A+B)\) to \(\dfrac{56}{65}\).

  8. Verify that \(A+B\) lies in the principal range of \(\sin^{-1}x\).

  9. Conclude the required identity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  30 steps
  1. Let
    \[A=\cos^{-1}\dfrac{12}{13}\]
  2. r
    \[\cos A=\dfrac{12}{13}\]
  3. Since
    \[0 < \dfrac{12}{13} < 1\]
    we have
    \[0 < A<\dfrac{\pi}{2}\]
    Hence, \(\sin A>0\)
  4. Using the identity
    \[\sin^2A+\cos^2A=1,\]
  5. we get
    \[ \sin^2A=1-\cos^2A\]
  6. Substituting \(\cos A=\dfrac{12}{13}\),
    \[\begin{aligned} \sin^2A&=1-\left(\dfrac{12}{13}\right)^2\\ &=1-\dfrac{144}{169}\\ &=\dfrac{169-144}{169}\\ &=\dfrac{25}{169} \end{aligned} \]
  7. Since \(A\) lies in the first quadrant, \(\sin A\) is positive. Therefore,
    \[\sin A=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}\]
  8. Thus,
    \[\boxed{\sin A=\dfrac{5}{13}}\]
  9. Now let
    \[B=\sin^{-1}\dfrac{3}{5}\]
  10. Therefore,
    \[\sin B=\dfrac{3}{5}\]
  11. Since
    \[0 < \dfrac{3}{5} < 1\]
    we have
    \[0 < B < \dfrac{\pi}{2}\]
    Hence, \(\cos B>0\).
  12. Using
    \[\sin^2B+\cos^2B=1\]
  13. we obtain
    \[\cos^2B=1-\sin^2B\]
  14. Substituting \(\sin B=\dfrac{3}{5}\),
    \[\begin{aligned} \cos^2B&=1-\left(\dfrac{3}{5}\right)^2\\ &=1-\dfrac{9}{25}\\ &=\dfrac{25-9}{25}\\ &=\dfrac{16}{25} \end{aligned} \]
  15. Since \(B\) lies in the first quadrant, \(\cos B\) is positive. Therefore,
    \[\begin{aligned}\cos B&=\sqrt{\dfrac{16}{25}}\\ &=\dfrac{4}{5}\end{aligned}\]
  16. Thus,
    \[\boxed{\cos B=\dfrac{4}{5}}\]
  17. Now consider
    \[A+B=\cos^{-1}\dfrac{12}{13}+\sin^{-1}\dfrac{3}{5}\]
  18. Using the sine addition formula,
    \[\sin(A+B)=\sin A\cos B+\cos A\sin B\]
  19. Substituting
    \[\sin A=\dfrac{5}{13},\quad\cos B=\dfrac{4}{5},\quad\cos A=\dfrac{12}{13},\quad\sin B=\dfrac{3}{5}\]
  20. we get
    \[\sin(A+B)=\left(\dfrac{5}{13}\right)\left(\dfrac{4}{5}\right)+\left(\dfrac{12}{13}\right)\left(\dfrac{3}{5}\right)\]
  21. Now simplify each term:
    \[ \left(\dfrac{5}{13}\right) \left(\dfrac{4}{5}\right) = \dfrac{20}{65}, \]
    \[ \left(\dfrac{12}{13}\right) \left(\dfrac{3}{5}\right) = \dfrac{36}{65}. \]
  22. Therefore,
    \[\begin{aligned}\sin(A+B)&=\dfrac{20}{65}+\dfrac{36}{65}\\ &=\dfrac{56}{65}\end{aligned}\]
  23. Hence,
    \[\sin(A+B)=\dfrac{56}{65}\]
  24. We now check the principal value condition. Since
    \[0 < A < \dfrac{\pi}{2}\]
    and
    \[0 < B < \dfrac{\pi}{2}\]
    we have
    \[0 < A+B < \pi\]
  25. Also,
    \[ \sin(A+B)=\dfrac{56}{65}>0. \]
    Thus, \(A+B\) could lie in either the first or second quadrant. We therefore need to establish that the sum is actually less than \(\dfrac{\pi}{2}\)
  26. Since
    \[A=\cos^{-1}\dfrac{12}{13}\]
    and
    \[B=\sin^{-1}\dfrac{3}{5}\]
    both are acute angles. Moreover,
    \[\begin{aligned}\tan A&=\dfrac{\sin A}{\cos A}\\ &=\dfrac{5/13}{12/13}\\ &=\dfrac{5}{12} \end{aligned} \]
    and
    \[\begin{aligned} \tan B&=\dfrac{\sin B}{\cos B}\\ &=\dfrac{3/5}{4/5}\\ &=\dfrac{3}{4} \end{aligned} \]
  27. Therefore,
    \[\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
  28. Substituting the values,
    \[\begin{aligned} \tan(A+B)&=\dfrac{\dfrac{5}{12}+\dfrac{3}{4}}\\ {1-\left(\dfrac{5}{12}\right)\left(\dfrac{3}{4}\right)}\\ &=\dfrac{\dfrac{5}{12}+\dfrac{9}{12}}{1-\dfrac{15}{48}}\\ &=\dfrac{\dfrac{14}{12}}{1-\dfrac{5}{16}}\\ &=\dfrac{\dfrac{7}{6}}{\dfrac{11}{16}}\\ &=\dfrac{7}{6}\times\dfrac{16}{11}\\ &=\dfrac{56}{33}>0 \end{aligned} \]
  29. Since \(0 < A+B < \pi\) and
    \[\tan(A+B)>0\]
    it follows that
    \[0 < A+B < \dfrac{\pi}{2}\]
  30. Therefore, \(A+B\) lies in the principal value range of \(\sin^{-1}x\):
    \[-\dfrac{\pi}{2}\leq A+B\leq\dfrac{\pi}{2}\]
  31. Hence, from
    \[\sin(A+B)=\dfrac{56}{65}\]
  32. we obtain
    \[A+B=\sin^{-1}\dfrac{56}{65}\]
  33. Substituting the values of \(A\) and \(B\),
    \[\boxed{\cos^{-1}\dfrac{12}{13}+\sin^{-1}\dfrac{3}{5}=\sin^{-1}\dfrac{56}{65}}\]
  34. Hence, Proved
🎯 Exam Significance
Exam Significance
›

This problem tests the use of the sine addition formula in conjunction with inverse trigonometric functions. For a complete board-examination proof, it is important to determine the correct signs of \(\sin A\) and \(\cos B\) from the principal ranges of the inverse functions.

The range verification at the end is also significant. Knowing \(\sin(A+B)=\dfrac{56}{65}\) alone does not automatically establish

\[ A+B=\sin^{-1}\dfrac{56}{65}, \]
because sine takes the same positive value at two angles in \([0,\pi]\). The principal-value condition must therefore be justified.

Significance for Competitive Entrance Examinations

This problem provides a useful pattern for competitive examinations: convert the given inverse trigonometric functions into ordinary trigonometric ratios and then apply an addition identity.

The Pythagorean triples

\[ 5,12,13 \]
and
\[ 3,4,5 \]
immediately give the required complementary ratios:

\[ \cos A=\dfrac{12}{13} \quad\Longrightarrow\quad \sin A=\dfrac{5}{13}, \]
\[ \sin B=\dfrac{3}{5} \quad\Longrightarrow\quad \cos B=\dfrac{4}{5}. \]

This recognition reduces the computational effort considerably and is especially useful in time-constrained entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. If \(A=\cos^{-1}\dfrac{12}{13}\), then \(\sin A=\dfrac{5}{13}\).

  2. If \(B=\sin^{-1}\dfrac{3}{5}\), then \(\cos B=\dfrac{4}{5}\).

  3. The sine addition formula is \(\sin(A+B)=\sin A\cos B+\cos A\sin B\).

  4. \(\sin(A+B)=\dfrac{20}{65}+\dfrac{36}{65}=\dfrac{56}{65}\).

  5. The intermediate product \(\dfrac{5}{13}\times\dfrac{4}{5}\) equals \(\dfrac{20}{65}\).

  6. The principal range of \(\sin^{-1}x\) is \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\).

  7. The range of \(A+B\) must be checked before replacing \(A+B\) by an inverse sine expression.

  8. The Pythagorean triples \(5,12,13\) and \(3,4,5\) make this problem particularly efficient to solve.
← Q5
6 / 14  ·  43%
Q7 →
Q7
NUMERIC3 marks
Prove that \(\tan^{-1}\dfrac{63}{16}=\sin^{-1}\dfrac{5}{13}+\cos^{-1}\dfrac{3}{5}\)
📘 Concept & Theory
Concept/Theory
›

This problem requires converting the given inverse sine and inverse cosine expressions into ordinary tangent ratios and then applying the tangent addition formula.

If

\[ A=\sin^{-1}x, \]
then
\[ \sin A=x. \]
Similarly, if
\[ B=\cos^{-1}x, \]
then
\[ \cos B=x. \]

Since both given values are positive, \(A\) and \(B\) lie in the first quadrant. Hence, the corresponding cosine and sine values are positive.

After finding \(\tan A\) and \(\tan B\), we use

\[ \tan(A+B) = \dfrac{\tan A+\tan B} {1-\tan A\tan B}. \]

Finally, the principal value condition must be checked. The principal value range of \(\tan^{-1}x\) is

\[ -\dfrac{\pi}{2}<\tan^{-1}x<\dfrac{\pi}{2}. \]

Therefore, after obtaining

\[ \tan(A+B)=\dfrac{63}{16}, \]
we must establish that \(A+B\) lies in the first quadrant before concluding that
\[ A+B=\tan^{-1}\dfrac{63}{16}. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(A=\sin^{-1}\dfrac{5}{13}\).

  2. Find \(\cos A\) and then \(\tan A\).

  3. Let \(B=\cos^{-1}\dfrac{3}{5}\).

  4. Find \(\sin B\) and then \(\tan B\).

  5. Apply the tangent addition formula to \(A+B\).

  6. Simplify the resulting value to \(\dfrac{63}{16}\).

  7. Verify that \(A+B\) lies in the principal range of \(\tan^{-1}x\).

  8. Conclude the required identity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  34 steps
  1. Let
    \[A=\sin^{-1}\dfrac{5}{13}\]
  2. Therefore,
    \[\sin A=\dfrac{5}{13}\]
  3. Since
    \[0 < \dfrac{5}{13} < 1\]
    we have
    \[0 < A < \dfrac{\pi}{2}\]
    Hence, \(\cos A>0\)
  4. Using the identity
    \[\sin^2A+\cos^2A=1\]
  5. we obtain
    \[\cos^2A=1-\sin^2A\]
  6. Substituting \(\sin A=\dfrac{5}{13}\),
    \[\begin{aligned}\cos^2A&=1-\left(\dfrac{5}{13}\right)^2\\ &=1-\dfrac{25}{169}\\ &=\dfrac{169-25}{169}\\ &=\dfrac{144}{169} \end{aligned} \]
  7. Since \(A\) lies in the first quadrant,
    \[\begin{aligned}\cos A&=\sqrt{\dfrac{144}{169}}\\&=\dfrac{12}{13}\end{aligned}\]
  8. Therefore,
    \[\begin{aligned} \tan A&=\dfrac{\sin A}{\cos A}\\ &=\dfrac{5/13}{12/13}\\ &=\dfrac{5}{12} \end{aligned} \]
  9. Thus,
    \[\boxed{\tan A=\dfrac{5}{12}}\]
  10. Now let
    \[B=\cos^{-1}\dfrac{3}{5}\]
  11. Therefore,
    \[\cos B=\dfrac{3}{5}\]
  12. Since
    \[0 < \dfrac{3}{5} < 1\]
    we have
    \[0 Hence, \(\sin B>0\).
  13. Using
    \[\sin^2B+\cos^2B=1\]
  14. we get
    \[\sin^2B=1-\cos^2B\]
  15. Substituting \(\cos B=\dfrac{3}{5}\),
    \[\begin{aligned} \sin^2B&=1-\left(\dfrac{3}{5}\right)^2\\ &=1-\dfrac{9}{25}\\ &=\dfrac{25-9}{25}\\ &=\dfrac{16}{25} \end{aligned} \]
  16. Since \(B\) lies in the first quadrant,
    \[\begin{aligned} \sin B&=\sqrt{\dfrac{16}{25}}\\ &=\dfrac{4}{5} \end{aligned} \]
  17. Therefore,
    \[\begin{aligned}\tan B&=\dfrac{\sin B}{\cos B}\\ &=\dfrac{4/5}{3/5}\\ &=\dfrac{4}{3} \end{aligned} \]
  18. Thus,
    \[\boxed{\tan B=\dfrac{4}{3}}\]
  19. Now apply the tangent addition formula:
    \[\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
  20. Substituting
    \[\tan A=\dfrac{5}{12}\quad\text{and}\]
    \[\tan B=\dfrac{4}{3}\]
  21. we get
    \[\tan(A+B)=\dfrac{\dfrac{5}{12}+\dfrac{4}{3}}{1-\left(\dfrac{5}{12}\right)\left(\dfrac{4}{3}\right)}\]
  22. Convert \(\dfrac{4}{3}\) into a fraction with denominator \(12\):
    \[\dfrac{4}{3}=\dfrac{16}{12}\]
  23. Therefore,
    \[\begin{aligned}\tan(A+B)&=\dfrac{\dfrac{5}{12}+\dfrac{16}{12}}{1-\dfrac{20}{36}}\\ &=\dfrac{\dfrac{21}{12}} {1-\dfrac{5}{9}} \end{aligned} \]
    Simplifying the denominator,
    \[\begin{aligned}1-\dfrac{5}{9}&=\dfrac{9-5}{9}\\ &=\dfrac{4}{9} \end{aligned} \]
  24. Hence,
    \[\tan(A+B)=\dfrac{21/12}{4/9}\]
  25. Dividing by \(\dfrac{4}{9}\),
    \[\begin{aligned} \tan(A+B)&=\dfrac{21}{12}\times\dfrac{9}{4}\ &=\dfrac{21\times9}{12\times4} \end{aligned} \]
  26. Since \(9/12=3/4\),
    \[\begin{aligned}\tan(A+B)&=\dfrac{21\times3}{4\times4}\\ &=\dfrac{63}{16} \end{aligned} \]
  27. Therefore,
    \[\tan(A+B)=\dfrac{63}{16}\]
  28. Verification of the Principal Value
  29. We have
    \[0 < A < \dfrac{\pi}{2}\quad\text{and}\]
    \[0 < B < \dfrac{\pi}{2}\]
  30. Therefore,
    \[0 < A+B < \pi\]
  31. Also,
    \[\tan(A+B)=\dfrac{63}{16}>0\]
  32. In the interval \((0,\pi)\), tangent is positive only in the first quadrant. Hence,
    \[0 < A+B < \dfrac{\pi}{2}\]
  33. Thus, \(A+B\) lies within the principal value range of \(\tan^{-1}x\):
    \[-\dfrac{\pi}{2}
  34. Consequently,
    \[A+B=\tan^{-1}\dfrac{63}{16}\]
  35. Substituting the definitions of \(A\) and \(B\),
    \[\boxed{\sin^{-1}\dfrac{5}{13}+\cos^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{63}{16}}\]
  36. Hence, Proved
🎯 Exam Significance
Exam Significance
›

This problem is important for board examinations because it combines inverse trigonometric functions with the tangent addition formula. It tests the student's ability to obtain missing trigonometric ratios and correctly handle principal values.

The final range verification is especially important. From

\[ \tan(A+B)=\dfrac{63}{16}, \]
it is not sufficient to immediately write
\[ A+B=\tan^{-1}\dfrac{63}{16}. \]
The location of \(A+B\) must be established so that the inverse tangent gives the correct angle.

Significance for Competitive Entrance Examinations

This problem is particularly efficient when solved by recognising the Pythagorean triples

\[ 5,12,13 \]
and
\[ 3,4,5. \]
They immediately give

\[ \sin A=\dfrac{5}{13} \quad\Longrightarrow\quad \tan A=\dfrac{5}{12}, \]
\[ \cos B=\dfrac{3}{5} \quad\Longrightarrow\quad \tan B=\dfrac{4}{3}. \]

The remaining calculation is then a direct application of the tangent addition formula. Recognising such numerical structures can save considerable time in JEE and other competitive entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. If \(A=\sin^{-1}\dfrac{5}{13}\), then \(\cos A=\dfrac{12}{13}\) and \(\tan A=\dfrac{5}{12}\).

  2. If \(B=\cos^{-1}\dfrac{3}{5}\), then \(\sin B=\dfrac{4}{5}\) and \(\tan B=\dfrac{4}{3}\).

  3. The tangent addition formula is \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\).

  4. \(\tan(A+B)=\dfrac{63}{16}\).

  5. Since \(A+B\in(0,\pi)\) and its tangent is positive, \(A+B\) lies in the first quadrant.

  6. Therefore, \(A+B\) belongs to the principal range of \(\tan^{-1}x\).

  7. The Pythagorean triples \(5,12,13\) and \(3,4,5\) provide a fast route to the required ratios.
← Q6
7 / 14  ·  50%
Q8 →
Q8
NUMERIC3 marks
Prove that \(\tan^{-1}\sqrt{x}=\dfrac{1}{2}\cos^{-1}\left(\dfrac{1-x}{1+x}\right),\quad x\in[0,1]\)
📘 Concept & Theory
Concept/Theory
›

This identity is based on the double-angle identity for cosine:

\[ \cos 2A = \dfrac{1-\tan^2A}{1+\tan^2A}. \]

The main idea is to introduce an angle \(A\) such that

\[ \tan A=\sqrt{x}. \]
Since \(x\in[0,1]\), we have
\[ 0\leq\sqrt{x}\leq1. \]
Therefore,
\[ 0\leq A\leq\dfrac{\pi}{4}. \]
Consequently,
\[ 0\leq2A\leq\dfrac{\pi}{2}, \]
which lies completely inside the principal value range of \(\cos^{-1}x\), namely \([0,\pi]\).

This range condition is essential because it allows us to use

\[ \cos^{-1}(\cos2A)=2A. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(A=\tan^{-1}\sqrt{x}\), so that \(\tan A=\sqrt{x}\).

  2. Square the relation to obtain \(\tan^2A=x\).

  3. Use the double-angle identity for cosine.

  4. Show that \(\dfrac{1-x}{1+x}=\cos2A\).

  5. Use the principal value range of \(\cos^{-1}\) to obtain \(\cos^{-1}(\cos2A)=2A\).

  6. Multiply by \(\dfrac12\) and substitute \(A=\tan^{-1}\sqrt{x}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  24 steps
  1. Let
    \[A=\tan^{-1}\sqrt{x}\]
  2. Therefore,
    \[\tan A=\sqrt{x}\]
  3. Squaring both sides,
    \[\tan^2A=x\]
  4. Since \(x\in[0,1]\)
    \[0\leq x\leq1\]
  5. Hence,
    \[0\leq\sqrt{x}\leq1\]
  6. Since \(A=\tan^{-1}\sqrt{x}\) and the principal value range of \(\tan^{-1}y\) is
    \[-\dfrac{\pi}{2}<\tan^{-1}y<\dfrac{\pi}{2},\]
    while \(\sqrt{x}\geq0\), we have
    \[0\leq A\leq\dfrac{\pi}{4}\]
  7. Therefore,
    \[0\leq2A\leq\dfrac{\pi}{2}\]
  8. Now consider the expression inside the inverse cosine:
    \[\dfrac{1-x}{1+x}\]
  9. Since \(x=\tan^2A\), we have
    \[\dfrac{1-x}{1+x}=\dfrac{1-\tan^2A}{1+\tan^2A}\]
  10. Using the identity
    \[1+\tan^2A=\sec^2A\]
  11. we get
    \[\dfrac{1-\tan^2A}{1+\tan^2A}=\dfrac{1-\tan^2A}{\sec^2A}\]
  12. Since
    \[\tan^2A=\sec^2A-1\]
  13. we obtain
    \[\dfrac{1-\tan^2A}{\sec^2A}=\dfrac{1-(\sec^2A-1)}{\sec^2A}\]
  14. Expanding the numerator,
    \[\begin{aligned}=\dfrac{1-\sec^2A+1}{\sec^2A}\\&=\dfrac{2-\sec^2A}{\sec^2A}\end{aligned}\]
  15. Separating the terms,
    \[=\dfrac{2}{\sec^2A}-1\]
  16. Since
    \[\dfrac{1}{\sec^2A}=\cos^2A\]
  17. we get
    \[= 2\cos^2A-1\]
  18. Using the double-angle identity
    \[\cos2A=2\cos^2A-1\]
  19. we obtain
    \[\dfrac{1-x}{1+x}=\cos2A\]
  20. Therefore, the right-hand side becomes
    \[\dfrac12\cos^{-1}\left(\dfrac{1-x}{1+x}\right)=\dfrac12\cos^{-1}(\cos2A)\]
  21. We have already established that
    \[0\leq2A\leq\dfrac{\pi}{2}\]
  22. Since this interval lies within the principal value range
    \[[0,\pi]\]
    of \(\cos^{-1}x\), we can write
    \[\cos^{-1}(\cos2A)=2A\]
  23. Hence,
    \[\dfrac12\cos^{-1}(\cos2A)=\dfrac12(2A)\]
    \[=A\]
  24. But
    \[ A=\tan^{-1}\sqrt{x}\]
  25. Therefore,
    \[\boxed{\tan^{-1}\sqrt{x}=\dfrac12\cos^{-1}\left(\dfrac{1-x}{1+x}\right)}\]
  26. for
    \[\boxed{x\in[0,1]}\]
  27. Hence, Proved
🎯 Exam Significance
Exam Significance
›

This is an important identity-proof question because it combines substitution, the Pythagorean identity, the double-angle formula, and the principal value concept of inverse trigonometric functions.

The condition

\[ x\in[0,1] \]
is essential. It guarantees that
\[ 0\leq A\leq\dfrac{\pi}{4}, \]
and consequently
\[ 0\leq2A\leq\dfrac{\pi}{2}. \]
Thus, the step
\[ \cos^{-1}(\cos2A)=2A \]
is valid.

Significance for Competitive Entrance Examinations

This identity is particularly useful in competitive examinations because it provides a direct conversion between inverse tangent and inverse cosine. Recognising the standard identity

\[ \cos2A = \dfrac{1-\tan^2A}{1+\tan^2A} \]

allows complicated-looking inverse trigonometric expressions to be reduced quickly.

The domain restriction should also be remembered. Without checking the principal range, the step involving \(\cos^{-1}(\cos2A)\) can produce an incorrect result.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Set \(A=\tan^{-1}\sqrt{x}\), so that \(\tan^2A=x\).

  2. Use \(\dfrac{1-\tan^2A}{1+\tan^2A}=\cos2A\).

  3. For \(x\in[0,1]\), \(0\leq A\leq\dfrac{\pi}{4}\).

  4. Therefore, \(0\leq2A\leq\dfrac{\pi}{2}\), which is within the principal range of \(\cos^{-1}\).

  5. Hence, \(\cos^{-1}(\cos2A)=2A\).

  6. Multiplication by \(\dfrac12\) gives the required left-hand side.

  7. The condition \(x\in[0,1]\) is essential for the stated identity.
← Q7
8 / 14  ·  57%
Q9 →
Q9
NUMERIC3 marks
Prove that \(\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\dfrac{x}{2},\quad x\in\left(0,\dfrac{\pi}{4}\right).\)
📘 Concept & Theory
Concept/Theory
›

This problem uses the half-angle identities

\[ 1+\sin x = \left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)^2 \]

and

\[ 1-\sin x = \left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)^2. \]

The given interval

\[ x\in\left(0,\dfrac{\pi}{4}\right) \]
is important because it ensures

\[ 0<\dfrac{x}{2}<\dfrac{\pi}{8}. \]

Therefore, both \(\sin\dfrac{x}{2}\) and \(\cos\dfrac{x}{2}\) are positive, and

\[ \cos\dfrac{x}{2}>\sin\dfrac{x}{2}. \]

Hence, when taking square roots, the positive signs can be selected:

\[ \sqrt{1+\sin x} = \sin\dfrac{x}{2}+\cos\dfrac{x}{2} \]

and

\[ \sqrt{1-\sin x} = \cos\dfrac{x}{2}-\sin\dfrac{x}{2}. \]

After substitution, the complicated fraction reduces directly to \(\cot\dfrac{x}{2}\). The final step uses the principal value property of \(\cot^{-1}x\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Use the given range of \(x\) to determine the signs of the half-angle expressions.

  2. Express \(1+\sin x\) as a perfect square involving \(\sin\dfrac{x}{2}\) and \(\cos\dfrac{x}{2}\).

  3. Express \(1-\sin x\) as another perfect square.

  4. Take the appropriate positive square roots.

  5. Substitute these expressions into the given fraction.

  6. Simplify the fraction to \(\cot\dfrac{x}{2}\).

  7. Apply \(\cot^{-1}(\cot\theta)=\theta\) using the principal value range.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  28 steps
  1. Given —
    \[x\in\left(0,\dfrac{\pi}{4}\right)\]
  2. Dividing the inequality by \(2\), we obtain
    \[0<\dfrac{x}{2}<\dfrac{\pi}{8}\]
  3. Therefore,
    \[\sin\dfrac{x}{2}>0,\quad\cos\dfrac{x}{2}>0,\]
    and, since \(0<\dfrac{x}{2}<\dfrac{\pi}{4}\),
    \[\cos\dfrac{x}{2}>\sin\dfrac{x}{2}\]
  4. Now consider
    \[1+\sin x\]
  5. sing the double-angle identity
    \[\sin x=2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\]
  6. we get
    \[1+\sin x=1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\]
  7. Also,
    \[1=\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}\]
  8. Therefore,
    \[1+\sin x=\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\]
  9. Using the identity
    \[a^2+b^2+2ab=(a+b)^2\]
  10. we obtain
    \[1+\sin x=\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)^2\]
  11. Hence,
    \[\sqrt{1+\sin x}=\sqrt{\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)^2}\]
  12. Since
    \[\sin\dfrac{x}{2}+\cos\dfrac{x}{2}>0,\]
    we have
    \[\boxed{\sqrt{1+\sin x}=\sin\dfrac{x}{2}+\cos\dfrac{x}{2}}\]
  13. Similarly, consider
    \[1-\sin x\]
  15. we obtain
    \[1-\sin x=1-2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\]
  16. Using
    \[1=\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}\]
  17. we get
    \[1-\sin x=\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}-2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\]
  18. Using
    \[a^2+b^2-2ab=(a-b)^2\]
  19. we obtain
    \[1-\sin x=\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)^2\]
  20. Therefore,
    \[\sqrt{1-\sin x}=\sqrt{\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)^2}\]
  21. Since
    \[\cos\dfrac{x}{2}>\sin\dfrac{x}{2},\]
    the quantity inside the square root is positive. Hence,
    \[\boxed{\sqrt{1-\sin x}=\cos\dfrac{x}{2}-\sin\dfrac{x}{2}}\]
  22. Now substitute these two results into the given expression:
    \[\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\]
  23. We get
    \[\cot^{-1}\left(\dfrac{\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)+\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)}{\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)-\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)}\right)\]
  24. Now simplify the numerator:
    \[\sin\dfrac{x}{2}+\cos\dfrac{x}{2}+\cos\dfrac{x}{2}-\sin\dfrac{x}{2}=2\cos\dfrac{x}{2}\]
  25. Similarly, the denominator becomes
    \[\sin\dfrac{x}{2}+\cos\dfrac{x}{2}-\cos\dfrac{x}{2}+\sin\dfrac{x}{2}=2\sin\dfrac{x}{2}\]
  26. Therefore, the expression becomes
    \[\cot^{-1}\left(\dfrac{2\cos\dfrac{x}{2}}{2\sin\dfrac{x}{2}}\right)\]
  27. Canceling \(2\),
    \[=\cot^{-1}\left(\dfrac{\cos\dfrac{x}{2}}{\sin\dfrac{x}{2}}\right)\]
  28. Since
    \[\cot\theta=\dfrac{\cos\theta}{\sin\theta},\]
  29. we obtain
    \[=\cot^{-1}\left(\cot\dfrac{x}{2}\right)\]
  30. The principal value range of \(\cot^{-1}y\) is
    \[0 < \cot^{-1} < \pi\]
  31. Since \(0 < \dfrac{x}{2} < \dfrac{\pi}{8} < \pi\) \(\dfrac{x}{2}\) lies within the principal value range of \(\cot^{-1}\). Therefore,
    \[\cot^{-1}\left(\cot\dfrac{x}{2}\right)=\dfrac{x}{2}\]
  32. Hence,
    \[\boxed{\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\dfrac{x}{2}}\]
  33. for
    \[\boxed{x\in\left(0,\dfrac{\pi}{4}\right)}\]
  34. Hence, Proved
🎯 Exam Significance
Exam Significance
›

This problem is important because it combines half-angle identities, square-root simplification, and the principal value of an inverse trigonometric function. The given restriction on \(x\) is not incidental: it determines the signs of the square roots and guarantees that the final inverse cotangent step is valid.

A complete board-examination solution should explicitly justify why

\[ \sqrt{(a)^2}=a \]
is valid in this problem. This is possible because the given interval ensures that the relevant expressions are positive.

Significance for Competitive Entrance Examinations

The key competitive-examination insight is to recognise that the apparently complicated radical expressions are perfect squares:

\[ 1+\sin x = \left( \sin\dfrac{x}{2}+\cos\dfrac{x}{2} \right)^2, \]
\[ 1-\sin x = \left( \cos\dfrac{x}{2}-\sin\dfrac{x}{2} \right)^2. \]

Once these are identified, the entire fraction collapses to \(\cot\dfrac{x}{2}\). Recognising this structure can make the problem much faster to solve in entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(1+\sin x=\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)^2\).

  2. \(1-\sin x=\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)^2\).

  3. The condition \(x\in\left(0,\dfrac{\pi}{4}\right)\) ensures all required signs are positive.

  4. \(\sqrt{1+\sin x}=\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\).

  5. \(\sqrt{1-\sin x}=\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\).

  6. The given fraction simplifies to \(\cot\dfrac{x}{2}\).

  7. The principal value range of \(\cot^{-1}y\) is \((0,\pi)\).

  8. Since \(0<\dfrac{x}{2}<\dfrac{\pi}{8}\), \(\cot^{-1}\left(\cot\dfrac{x}{2}\right)=\dfrac{x}{2}\).
← Q8
9 / 14  ·  64%
Q10 →
Q10
NUMERIC3 marks
Prove that \( \tan^{-1}\left( \dfrac{ \sqrt{1+x}-\sqrt{1-x} }{ \sqrt{1+x}+\sqrt{1-x} } \right) = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}x, \qquad -\dfrac{1}{\sqrt{2}}\leq x\leq1\)
📘 Concept & Theory
Concept/Theory
›

This problem combines the properties of inverse trigonometric functions, half-angle identities, and the principal value ranges of \(\cos^{-1}x\) and \(\tan^{-1}x\).

The central idea is to express \(x\) in the form

\[ x=\cos 2\theta. \]

Since

\[ 1+\cos2\theta=2\cos^2\theta \]
and
\[ 1-\cos2\theta=2\sin^2\theta, \]
the radical expressions can be simplified in terms of \(\sin\theta\) and \(\cos\theta\).

The resulting fraction is

\[ \dfrac{\cos\theta-\sin\theta} {\cos\theta+\sin\theta}, \]

which can be recognised using the tangent subtraction identity:

\[ \tan\left(\dfrac{\pi}{4}-\theta\right) = \dfrac{1-\tan\theta}{1+\tan\theta} = \dfrac{\cos\theta-\sin\theta} {\cos\theta+\sin\theta}. \]

Therefore, the left-hand side becomes

\[ \tan^{-1} \left[ \tan\left(\dfrac{\pi}{4}-\theta\right) \right]. \]

The given range of \(x\) is essential because it ensures that \(\dfrac{\pi}{4}-\theta\) lies within the principal value range of \(\tan^{-1}\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Put \(x=\cos2\theta\).

  2. Determine the range of \(\theta\) from the given range of \(x\).

  3. Use \(1+\cos2\theta=2\cos^2\theta\) and \(1-\cos2\theta=2\sin^2\theta\).

  4. Take the correct positive square roots.

  5. Substitute the radical expressions into the given fraction.

  6. Rationalise the fraction to obtain an expression involving \(\tan\theta\).

  7. Recognise the result as \(\tan\left(\dfrac{\pi}{4}-\theta\right)\).

  8. Apply the principal value property of \(\tan^{-1}\).

  9. Use \(2\theta=\cos^{-1}x\) to obtain the required result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  46 steps
  1. Let
    \[x=\cos2\theta\]
  2. Since
    \[-\dfrac{1}{\sqrt{2}}\leq x\leq1\]
  3. we have
    \[-\dfrac{1}{\sqrt{2}}\leq\cos2\theta\leq1\]
  4. We choose
    \[2\theta=\cos^{-1}x\]
  5. Since
    \[-\dfrac{1}{\sqrt{2}}\leq x\leq1,\]
    , the principal value of \(\cos^{-1}x\) satisfies
    \[0\leq\cos^{-1}x\leq\dfrac{3\pi}{4}\]
  6. Therefore,
    \[0\leq2\theta\leq\dfrac{3\pi}{4}\]
  7. Dividing by \(2\),
    \[0\leq\theta\leq\dfrac{3\pi}{8}\]
  8. Hence, both \(\sin\theta\) and \(\cos\theta\) are non-negative. Therefore, the positive square roots can be used in the following steps.
  9. Step 1: Simplify \(\sqrt{1+x}\)
  10. Since
    \[x=\cos2\theta\]
  11. we have
    \[1+x=1+\cos2\theta\]
  12. Using the identity
    \[1+\cos2\theta=2\cos^2\theta\]
  13. we obtain
    \[1+x=2\cos^2\theta\]
  14. Taking the square root,
    \[\sqrt{1+x}=\sqrt{2\cos^2\theta}\]
  15. Since \(\cos\theta\geq0\),
    \[\boxed{\sqrt{1+x}=\sqrt{2}\cos\theta}\]
  16. Step 2: Simplify \(\sqrt{1-x}\)
  17. Similarly,
    \[1-x=1-\cos2\theta\]
  18. Using
    \[1-\cos2\theta=2\sin^2\theta\]
  19. we obtain
    \[1-x=2\sin^2\theta\]
  20. Taking the square root,
    \[\sqrt{1-x}=\sqrt{2\sin^2\theta}\]
  21. Since \(\sin\theta\geq0\),
    \[\boxed{\sqrt{1-x}=\sqrt{2}\sin\theta}\]
  22. Step 3: Substitute in the Given Fraction
  23. Consider
    \[E=\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\]
  24. Substituting the expressions obtained above,
    \[E=\dfrac{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}\]
  25. Taking \(\sqrt{2}\) common from the numerator and denominator,
    \[E=\dfrac{\sqrt{2}(\cos\theta-\sin\theta)}{\sqrt{2}(\cos\theta+\sin\theta)}\]
  26. Canceling \(\sqrt{2}\),
    \[E=\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\]
  27. Step 4: Convert the Fraction into a Tangent Form
  28. Multiply the numerator and denominator by \(\cos\theta\):
    \[E=\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\cdot\dfrac{\cos\theta}{\cos\theta}.\]
  29. Thus,
    \[E=\dfrac{\cos^2\theta-\sin\theta\cos\theta}{\cos^2\theta+\sin\theta\cos\theta}\]
  30. A more direct way is to divide the numerator and denominator by \(\cos\theta\). Since
    \[ \cos\theta>0 \]
    in the relevant interval, this is valid:
    \[E=\dfrac{1-\tan\theta}{1+\tan\theta}\]
  31. Using the tangent subtraction identity
    \[\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\]
  32. put
    \[A=\dfrac{\pi}{4},\quad B=\theta\]
  33. Since
    \[\tan\dfrac{\pi}{4}=1\]
  34. we obtain
    \[\tan\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{1-\tan\theta}{1+\tan\theta}\]
  35. Therefore,
    \[\boxed{E=\tan\left(\dfrac{\pi}{4}-\theta\right)}\]
  36. Step 5: Apply the Inverse Tangent
  37. The original left-hand side is
    \[\tan^{-1}(E)\]
  38. Therefore,
    \[\tan^{-1}(E)=\tan^{-1}\left[\tan\left(\dfrac{\pi}{4}-\theta\right)\right]\]
  39. From
    \[0\leq\theta\leq\dfrac{3\pi}{8}\]
  40. we get
    \[-\dfrac{\pi}{8}\leq\dfrac{\pi}{4}-\theta\leq\dfrac{\pi}{4}\]
  41. Hence,
    \[-\dfrac{\pi}{2} < \dfrac{\pi}{4}-\theta < \dfrac{\pi}{2}\]
  42. Thus, \(\dfrac{\pi}{4}-\theta\) lies within the principal value range of \(\tan^{-1}\). Therefore,
    \[\tan^{-1}\left[\tan\left(\dfrac{\pi}{4}-\theta\right)\right]=\dfrac{\pi}{4}-\theta\]
  43. Hence,
    \[\boxed{\tan^{-1}\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\dfrac{\pi}{4}-\theta}\]
  44. Step 6: Express \(\theta\) in Terms of \(x\)
  45. We started with
    \[x=\cos2\theta\]
  46. Therefore,
    \[2\theta=\cos^{-1}x\]
  47. Hence,
    \[\theta=\dfrac{1}{2}\cos^{-1}x\]
  48. Substituting this value of \(\theta\),
    \[\tan^{-1}\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x\]
  49. Therefore,
    \[\boxed{\tan^{-1}\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x}\]
    for
    \[-\dfrac{1}{\sqrt{2}}\leq x\leq1\]
  50. Hence, Proved
🎯 Exam Significance
Exam Significance
›

This problem is important for CBSE Class 12 Mathematics because it combines several frequently tested concepts: principal values, half-angle identities, square-root simplification, and inverse trigonometric transformations.

Particular attention should be given to the signs while evaluating

\[ \sqrt{2\cos^2\theta} \]
and
\[ \sqrt{2\sin^2\theta}. \]
The given range of \(x\) determines the corresponding range of \(\theta\), allowing the correct positive square roots to be selected.

The final principal-value check is equally important. It establishes that

\[ \tan^{-1}(\tan y)=y \]
can be applied to
\[ y=\dfrac{\pi}{4}-\theta. \]

Significance for Competitive Entrance Examinations

The main competitive-examination insight is to recognise the standard pattern

\[ \dfrac{\cos\theta-\sin\theta} {\cos\theta+\sin\theta} = \tan\left(\dfrac{\pi}{4}-\theta\right). \]

Once \(x=\cos2\theta\) is introduced, the two radicals immediately become half-angle expressions. This transforms a complicated radical expression into a simple inverse-tangent form.

The range

\[ -\dfrac{1}{\sqrt{2}}\leq x\leq1 \]
is also crucial because it guarantees that the resulting angle remains inside the principal value interval of \(\tan^{-1}\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  10 points

  1. Use \(x=\cos2\theta\) when expressions contain both \(1+x\) and \(1-x\).

  2. \(1+\cos2\theta=2\cos^2\theta\).

  3. \(1-\cos2\theta=2\sin^2\theta\).

  4. For the relevant range, \(\sqrt{1+x}=\sqrt{2}\cos\theta\).

  5. For the relevant range, \(\sqrt{1-x}=\sqrt{2}\sin\theta\).

  6. \(\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\tan\left(\dfrac{\pi}{4}-\theta\right)\).

  7. The given range implies \(0\leq\theta\leq\dfrac{3\pi}{8}\).

  8. Therefore, \(-\dfrac{\pi}{8}\leq\dfrac{\pi}{4}-\theta\leq\dfrac{\pi}{4}\), which lies within the principal range of \(\tan^{-1}\).

  9. Since \(2\theta=\cos^{-1}x\), we have \(\theta=\dfrac{1}{2}\cos^{-1}x\).

  10. The final result is \(\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x\).
← Q9
10 / 14  ·  71%
Q11 →
Q11
NUMERIC3 marks
Solve \(2\tan^{-1}(\cos x)=\tan^{-1}(2\text{cosec }x)\)
📘 Concept & Theory
Concept/Theory
›

This equation involves inverse tangent functions on both sides. The main idea is to introduce

\[ A=\tan^{-1}(\cos x) \]
and use the double-angle formula for tangent:

\[ \tan 2A = \dfrac{2\tan A}{1-\tan^2A}. \]

Since

\[ \tan A=\cos x, \]
this gives

\[ \tan 2A = \dfrac{2\cos x}{1-\cos^2x}. \]

Using

\[ 1-\cos^2x=\sin^2x, \]

we obtain

\[ \tan 2A = \dfrac{2\cos x}{\sin^2x}. \]

On the other hand,

\[ \tan\left(\tan^{-1}(2\text{cosec }x)\right) = 2\text{cosec }x = \dfrac{2}{\sin x}. \]

Equating these expressions will lead to

\[ \cos x=\sin x. \]
However, because inverse tangent is a principal-value function, we must verify that the resulting values of \(x\) actually satisfy the original equation.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. First note the domain restriction \(\sin x\neq0\).

  2. Put \(A=\tan^{-1}(\cos x)\).

  3. Use the double-angle formula to calculate \(\tan 2A\).

  4. Express \(\tan 2A\) in terms of \(\sin x\) and \(\cos x\).

  5. Take the tangent of the given equation.

  6. Equate the two tangent values and simplify.

  7. Obtain \(\sin x=\cos x\).

  8. Find the general values of \(x\).

  9. Verify that these values satisfy the original inverse-trigonometric equation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  22 steps
  1. Given — \(2\tan^{-1}(\cos x)=\tan^{-1}(2\text{cosec }x)\)
  2. Step 1: Determine the Domain
  3. Since \(\text{cosec }x\) occurs in the equation, it must be defined. Therefore,
    \[\sin x\neq0\]
  4. Hence,
    \[x\neq n\pi,\quad n\in\mathbb Z\]
  5. Step 2: Introduce an Auxiliary Angle
  6. Let
    \[A=\tan^{-1}(\cos x)\]
  7. Therefore,
    \[\tan A=\cos x\]
  8. Since
    \[ -1\leq\cos x\leq1, \]
    and the principal value range of \(\tan^{-1}y\) is
    \[-\dfrac{\pi}{2}<\tan^{-1}y<\dfrac{\pi}{2}\]
  9. we have
    \[-\dfrac{\pi}{4}\leq A\leq\dfrac{\pi}{4}\]
  10. Consequently,
    \[-\dfrac{\pi}{2}\leq2A\leq\dfrac{\pi}{2}\]
  11. Thus, \(2A\) lies within the principal value range of \(\tan^{-1}\), apart from the fact that the endpoints require separate attention. This will allow us to compare the tangent values carefully.
  12. Step 3: Find \(\tan 2A\)
  13. Using the double-angle formula,
    \[\tan 2A=\dfrac{2\tan A}{1-\tan^2A}\]
  14. Since
    \[\tan A=\cos x\]
  15. we get
    \[\tan 2A=\dfrac{2\cos x}{1-\cos^2x}\]
  16. Using
    \[1-\cos^2x=\sin^2x\]
  17. we obtain
    \[\tan 2A=\dfrac{2\cos x}{\sin^2x}\]
  18. Step 4: Take Tangent on Both Sides
  19. The given equation is
    \[2A=\tan^{-1}(2\text{cosec }x)\]
  20. Taking tangent on both sides,
    \[\tan 2A=\tan\left(\tan^{-1}(2\text{cosec }x)\right)\]
  21. Therefore,
    \[\tan 2A=2\text{cosec }x\]
  22. Since
    \[\text{cosec }x=\dfrac{1}{\sin x}\]
  23. we get
    \[\tan 2A=\dfrac{2}{\sin x}\]
  24. But we have already obtained
    \[\tan 2A=\dfrac{2\cos x}{\sin^2x}\]
  25. Hence,
    \[\dfrac{2\cos x}{\sin^2x}=\dfrac{2}{\sin x}\]
  26. Since \(\sin x\neq0\), we may multiply both sides by \(\sin^2x\):
    \[2\cos x=2\sin x\]
  27. Dividing both sides by \(2\),
    \[\cos x=\sin x\]
  28. Therefore,
    \[\tan x=1\]
  29. Step 5: Find the General Solution
  30. The general solution of \9\tan x=1\) is
    \[x=\dfrac{\pi}{4}+n\pi, \quad n\in\mathbb Z\]
🎯 Exam Significance
Exam Significance
›

This problem is important because it tests the correct handling of inverse tangent functions, the double-angle formula for tangent, and the domain of \(\text{cosec }x\).

The key step is to recognise that the equation can be transformed using

\[ \tan 2A = \dfrac{2\tan A}{1-\tan^2A}. \]
However, a complete solution should not stop after obtaining
\[ \tan x=1. \]
The resulting values must be checked against the original equation because inverse trigonometric functions have restricted principal-value ranges.

Significance for Competitive Entrance Examinations

For competitive examinations, the efficient route is to introduce

\[ A=\tan^{-1}(\cos x) \]
and compare the tangent of both sides. This immediately reduces the inverse-trigonometric equation to the elementary condition

\[ \cos x=\sin x. \]

The important competitive-examination habit is to verify the resulting general solution in the original equation rather than relying solely on the tangent equation.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. The equation is defined only when \(\sin x\neq0\).

  2. Set \(A=\tan^{-1}(\cos x)\), so that \(\tan A=\cos x\).

  3. Use \(\tan2A=\dfrac{2\tan A}{1-\tan^2A}\).

  4. This gives \(\tan2A=\dfrac{2\cos x}{\sin^2x}\).

  5. The right-hand side of the original equation has tangent \(\dfrac{2}{\sin x}\).

  6. Equating the two gives \(\cos x=\sin x\).

  7. Therefore, \(\tan x=1\).

  8. The general solution is \(x=\dfrac{\pi}{4}+n\pi\), \(n\in\mathbb Z\).

  9. The solutions satisfy the original inverse-trigonometric equation for both even and odd \(n\).
← Q10
11 / 14  ·  79%
Q12 →
Q12
NUMERIC3 marks
Solve \(\tan^{-1}\left(\dfrac{1+x}{1-x}\right)=\dfrac{1}{2}\tan^{-1}(x),\quad x>0.\)
📘 Concept & Theory
Concept/Theory
›

This equation contains an inverse tangent expression and a half-angle involving \(\tan^{-1}x\). The natural approach is to introduce

\[ A=\dfrac{1}{2}\tan^{-1}x. \]

Then

\[ 2A=\tan^{-1}x, \]

and therefore

\[ \tan 2A=x. \]

We can then compare this with the tangent of the left-hand side. However, the domain must be considered carefully because the expression

\[ \dfrac{1+x}{1-x} \]
is undefined at \(x=1\), and its sign changes when \(x\) crosses \(1\).

Since \(x>0\), the right-hand side is always positive:

\[ \dfrac12\tan^{-1}x>0. \]

For \(x>1\), however,

\[ \dfrac{1+x}{1-x}<0, \]
so the left-hand side is negative. Hence \(x>1\) cannot give a solution.

It therefore remains to investigate \(0 < x < 1\). In this interval both sides are positive, so the tangent function can be used without crossing a principal-value ambiguity.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the domain restriction \(x\neq1\).

  2. Use the sign of the two sides to eliminate \(x>1\).

  3. Consider \(0

  4. Set \(A=\dfrac12\tan^{-1}x\).

  5. Use the double-angle formula for \(\tan 2A\).

  6. Take tangent on both sides of the original equation.

  7. Equate the resulting expressions.

  8. Show that the resulting algebraic equation has no real solution.

  9. Conclude that the given equation has no real solution for \(x>0\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  43 steps
  1. Given — \(\tan^{-1}\left(\dfrac{1+x}{1-x}\right)=\dfrac12\tan^{-1}x,\quad x>0\)
  2. Step 1: Domain of the Equation
  3. The expression
    \[\dfrac{1+x}{1-x}\]
    is defined only when
    \[1-x\neq0\]
  4. Therefore,
    \[x\neq1\]
  5. Since \(x>0\), there are two possible intervals to consider:
    \[0 < x < 1\quad\text{and}\quad x>1\]
  6. Step 2: Eliminate \(x>1\)
  7. Suppose
    \[x>1\]
    then
    \[1+x>0\quad\text{and}\quad 1-x < 0\]
  8. Hence,
    \[\dfrac{1+x}{1-x} < 0\]
  9. Therefore,
    \[\tan^{-1}\left(\dfrac{1+x}{1-x}\right) < 0\]
  10. On the other hand, since \(x>0\),
    \[\tan^{-1}x>0\]
  11. Consequently,
    \[\dfrac12\tan^{-1}x>0\]
  12. Thus, for \(x>1\), the left-hand side is negative while the right-hand side is positive. Hence, no solution is possible for \(x>1\).
  13. Therefore, any possible solution must satisfy
    \[0 < x < 1\]
  14. Step 3: Introduce an Auxiliary Angle
  15. Let
    \[A=\dfrac12\tan^{-1}x\]
  16. Therefore,
    \[2A=\tan^{-1}x\]
  17. Taking tangent on both sides,
    \[\tan2A=x\]
  18. Since \(0 < x < 1\), we have
    \[0 < \tan^{-1}x < \dfrac{\pi}{4}\]
  19. Hence,
    \[0 < A < \dfrac{\pi}{8}\]
  20. and therefore
    \[0 < 2A < \dfrac{\pi}{4}\]
  21. Step 4: Take Tangent on Both Sides
  22. The original equation is
    \[\tan^{-1}\left(\dfrac{1+x}{1-x}\right)=A\]
  23. Since \(0 < x < 1\),
    \[\dfrac{1+x}{1-x}>0\]
  24. Therefore,
    \[0 < \tan^{-1}\left(\dfrac{1+x}{1-x}\right) < \dfrac{\pi}{2}\]
  25. Also,
    \[0 < A < \dfrac{\pi}{8}\]
  26. Thus both sides lie in an interval where the tangent function is one-to-one. Taking tangent on both sides is therefore valid.
  27. We obtain
    \[\tan\left[\tan^{-1}\left(\dfrac{1+x}{1-x}\right)\right]=\tan A\]
  28. Hence,
    \[\dfrac{1+x}{1-x}=\tan A\]
  29. Step 5: Use the Double-Angle Relation
  30. We already have
    \[\tan2A=x\]
  31. Using the double-angle formula,
  32. \[\tan2A=\dfrac{2\tan A}{1-\tan^2A}\]
  33. Therefore,
    \[x=\dfrac{2\tan A}{1-\tan^2A}\]
  34. Let
    \[t=\tan A\]
    then
    \[x=\dfrac{2t}{1-t^2}\]
  35. From Step 4,
    \[t=\dfrac{1+x}{1-x}\]
  36. Substituting this into the double-angle relation gives
    \[x=\dfrac{2\left(\dfrac{1+x}{1-x}\right)}{1-\left(\dfrac{1+x}{1-x}\right)^2}\]
  37. Now simplify the denominator:
    \[1-\left(\dfrac{1+x}{1-x}\right)^2=\dfrac{(1-x)^2-(1+x)^2}{(1-x)^2}\]
  38. Expanding both squares,
    \[(1-x)^2=1-2x+x^2\]
    and
    \[(1+x)^2=1+2x+x^2\]
  39. Therefore,
    \[\begin{aligned}(1-x)^2-(1+x)^2&=(1-2x+x^2)-(1+2x+x^2)\\ &=-4x\end{aligned}\]
  40. Hence,
    \[1-\left(\dfrac{1+x}{1-x}\right)^2=\dfrac{-4x}{(1-x)^2}\]
  41. Also,
    \[2\left(\dfrac{1+x}{1-x}\right)=\dfrac{2(1+x)}{1-x}\]
  42. Therefore,
    \[x=\dfrac{\dfrac{2(1+x)}{1-x}}{\dfrac{-4x}{(1-x)^2}}\]
  43. Dividing by a fraction is equivalent to multiplying by its reciprocal:
    \[x=\dfrac{2(1+x)}{1-x}\times\dfrac{(1-x)^2}{-4x}\]
  44. Canceling one factor of \(1-x\),
    \[x=-\dfrac{(1+x)(1-x)}{2x}\]
  45. Using
    \[(1+x)(1-x)=1-x^2\]
  46. we get
    \[x=-\dfrac{1-x^2}{2x}\]
  47. Multiplying both sides by \(2x\), which is positive because \(x>0\),
    \[2x^2=-(1-x^2)\]
  48. Therefore,
    \[2x^2=-1+x^2\]
  49. This has no real solution because
  50. \[x^2\geq0\]
    for every real \(x\), whereas
    \[-1<0\]
  51. Thus, there is no real value of \(x\) satisfying the equation.
🎯 Exam Significance
Exam Significance
›

This problem is important because it tests the interaction between inverse trigonometric functions, principal values, and the tangent double-angle identity.

A common mistake is to manipulate the equation algebraically without first considering the sign of

\[ \dfrac{1+x}{1-x}. \]
For \(x>1\), the left-hand side is negative while the right-hand side is positive, immediately ruling out that interval.

For the remaining interval \(0 < x < 1\), taking tangent is safe because both sides lie in an interval where tangent is one-to-one. The resulting contradiction

\[ x^2=-1 \]
establishes that no real solution exists.

Significance for Competitive Entrance Examinations

The fastest conceptual observation is the sign analysis:

\[ x>1 \quad\Longrightarrow\quad \dfrac{1+x}{1-x}<0, \]

whereas

\[ x>0 \quad\Longrightarrow\quad \dfrac12\tan^{-1}x>0. \]

Hence, \(x>1\) is immediately impossible. For \(0 < x < 1\), the double-angle formula reduces the equation to the impossible condition

\[ x^2=-1. \]

This illustrates an important competitive-examination strategy: use domain and sign information before performing lengthy inverse-trigonometric manipulations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. The expression \(\dfrac{1+x}{1-x}\) requires \(x\neq1\).

  2. For \(x>1\), the left-hand side is negative while the right-hand side is positive.

  3. Therefore, any possible solution must satisfy \(0 < x < 1\).

  4. Set \(A=\dfrac12\tan^{-1}x\), so that \(\tan2A=x\).

  5. For \(0

  6. The resulting equation leads to \(x^2=-1\).

  7. Since \(x^2=-1\) has no real solution, the original equation has no real solution.

  8. The final answer is \(\boxed{\text{No real solution}}\).
← Q11
12 / 14  ·  86%
Q13 →
Q13
NUMERIC3 marks
Evaluate \(\sin\left(\tan^{-1}x\right), \quad |x| < 1\) Choose the correct option:
\( \begin{aligned} \text{A)}\quad &\dfrac{x}{\sqrt{1-x^2}}\\ \text{B)}\quad &\dfrac{1}{\sqrt{1-x^2}}\\ \text{C)}\quad &\dfrac{1}{\sqrt{1+x^2}}\\ \text{D)}\quad &\dfrac{x}{\sqrt{1+x^2}} \end{aligned} \)
📘 Concept & Theory
Concept/Theory
›

When an expression contains

\[ \sin(\tan^{-1}x), \]
the most direct method is to introduce an angle
\[ \theta=\tan^{-1}x. \]
Then
\[ \tan\theta=x. \]

We can represent \(\tan\theta\) as a ratio of the opposite and adjacent sides of a right triangle:

\[ \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=x. \]

Taking the adjacent side as \(1\), the opposite side is \(x\). By the Pythagorean theorem, the hypotenuse is

\[ \sqrt{1+x^2}. \]

Therefore,

\[ \sin\theta = \dfrac{x}{\sqrt{1+x^2}}. \]

The condition

\[|x| < 1\]
ensures, in particular, that \(x\) is finite and
\[ 1+x^2>0, \]
so the denominator is always real and non-zero.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Put \(\theta=\tan^{-1}x\).

  2. Convert the inverse tangent relation into \(\tan\theta=x\).

  3. Use the identity \(1+\tan^2\theta=\sec^2\theta\) to obtain \(\cos\theta\).

  4. Use \(\sin\theta=\tan\theta\cos\theta\).

  5. Substitute \(\tan\theta=x\) and simplify.

  6. Compare the result with the given options.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  13 steps
  1. Let
    \[\theta=\tan^{-1}x\]
  2. Therefore,
    \[\tan\theta=x\]
  3. Using the identity
    \[1+\tan^2\theta=\sec^2\theta\]
  4. we get
    \[1+x^2=\sec^2\theta\]
  5. Taking the positive square root,
    \[\sec\theta=\sqrt{1+x^2}\]
  6. We take the positive square root because
    \[\theta=\tan^{-1}x\]
    lies in the principal value range
    \[\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}\]
    where \(\cos\theta>0\) and hence
    \[\sec\theta>0\]
  7. Therefore,
    \[\cos\theta=\dfrac{1}{\sec\theta}=\dfrac{1}{\sqrt{1+x^2}}\]
  8. Now use
    \[\tan\theta=\dfrac{\sin\theta}{\cos\theta}\]
  9. Therefore,
    \[\sin\theta=\tan\theta\cos\theta\]
  10. Substituting \(\tan\theta=x\) and \(\cos\theta=\dfrac{1}{\sqrt{1+x^2}}\)
    we obtain
    \[\sin\theta=x\left(\dfrac{1}{\sqrt{1+x^2}}\right)\]
  11. Hence,
    \[\sin\theta=\dfrac{x}{\sqrt{1+x^2}}\]
  12. Since
    \[\theta=\tan^{-1}x\]
  13. we finally get
    \[\boxed{\sin\left(\tan^{-1}x\right)=\dfrac{x}{\sqrt{1+x^2}}}\]
  14. Therefore, the correct option is
    \[\boxed{\text{D) }\dfrac{x}{\sqrt{1+x^2}}}\]
🎯 Exam Significance
Exam Significance
›

This is a standard inverse-trigonometric transformation that tests whether the student can convert an inverse function into an ordinary trigonometric ratio.

The essential substitution is

\[ \theta=\tan^{-1}x. \]

This immediately gives

\[ \tan\theta=x. \]

From there, either the Pythagorean identity or a right-triangle construction gives the required sine value.

For board examinations, remembering the standard results

\[ \sin(\tan^{-1}x) = \dfrac{x}{\sqrt{1+x^2}}, \]
\[ \cos(\tan^{-1}x) = \dfrac{1}{\sqrt{1+x^2}}, \]
\[ \tan(\sin^{-1}x) = \dfrac{x}{\sqrt{1-x^2}} \]

can significantly reduce calculation time.

Significance for Competitive Entrance Examinations

This is a frequently useful transformation in JEE and other competitive examinations. The expression

\[ \sin(\tan^{-1}x) \]
should immediately suggest the substitution
\[ \theta=\tan^{-1}x. \]

The resulting right triangle has side ratio

\[ x:1:\sqrt{1+x^2}, \]
making the answer almost immediate.

A useful pattern to remember is that expressions involving \(\tan^{-1}x\) naturally produce

\[ \sqrt{1+x^2}, \]
whereas expressions involving \(\sin^{-1}x\) naturally produce
\[ \sqrt{1-x^2}. \]

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Set \(\theta=\tan^{-1}x\).

  2. Then \(\tan\theta=x\).

  3. Using \(1+\tan^2\theta=\sec^2\theta\), we get \(\sec\theta=\sqrt{1+x^2}\).

  4. Therefore, \(\cos\theta=\dfrac{1}{\sqrt{1+x^2}}\).

  5. Using \(\sin\theta=\tan\theta\cos\theta\), we obtain \(\sin\theta=\dfrac{x}{\sqrt{1+x^2}}\).

  6. The correct answer is option D.

  7. The restriction \(|x| < 1\) is compatible with the expression and ensures \(1+x^2>0\).
← Q12
13 / 14  ·  93%
Q14 →
Q14
NUMERIC3 marks
Solve \(\sin^{-1}(1-x)-2\sin^{-1}x=\dfrac{\pi}{2}\)

Choose the correct option:

\[ \begin{aligned} \text{A)}\quad &0,\dfrac{1}{2}\\ \text{B)}\quad &1,\dfrac{1}{2}\\ \text{C)}\quad &0\\ \text{D)}\quad &\dfrac{1}{2} \end{aligned} \]
📘 Concept & Theory
Concept/Theory
›

The key concept in this question is the principal value range of the inverse sine function.

For any real number \(y\) in the domain of \(\sin^{-1}y\),

\[ -\dfrac{\pi}{2}\leq\sin^{-1}y\leq\dfrac{\pi}{2}. \]

Therefore, the left-hand side contains a term

\[ \sin^{-1}(1-x) \]
whose value can never exceed \(\dfrac{\pi}{2}\).

Since \(\sin^{-1}x\geq0\) whenever \(x\geq0\), the equation

\[ \sin^{-1}(1-x)-2\sin^{-1}x=\dfrac{\pi}{2} \]
can attain \(\dfrac{\pi}{2}\) only when
\[ \sin^{-1}x=0. \]
This immediately gives
\[ x=0. \]

We should also check the domain. Both inverse sine expressions must be defined, so

\[ -1\leq x\leq1 \]

and

\[ -1\leq1-x\leq1. \]

The second condition gives

\[ 0\leq x\leq2. \]

Combining these conditions,

\[ 0\leq x\leq1. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Determine the domain of the inverse sine expressions.

  2. Use the principal value range of \(\sin^{-1}x\).

  3. Observe that \(2\sin^{-1}x\geq0\) for \(0\leq x\leq1\).

  4. Since \(\sin^{-1}(1-x)\leq\dfrac{\pi}{2}\), the equation can equal \(\dfrac{\pi}{2}\) only if \(2\sin^{-1}x=0\).

  5. Obtain \(x=0\).

  6. Verify \(x=0\) in the original equation.

  7. Select the correct option.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  19 steps
  1. Given — \(\sin^{-1}(1-x)-2\sin^{-1}x=\dfrac{\pi}{2}\)
  2. Step 1: Determine the Domain
  3. Since \(\sin^{-1}x\) is defined only when
    \[-1\leq x\leq1,\]
    we require
    \[-1\leq x\leq1\]
    Also, for \(\sin^{-1}(1-x)\) to be defined,
    \[-1\leq1-x\leq1\]
  4. Solving the first inequality,
    \[\begin{aligned}-1&\leq1-x\\x&\leq2\end{aligned}\]
  5. Solving the second inequality,
    \[1-x\leq1\]
    \[-x\leq0\]
    \[-x\leq0\]
  6. Therefore,
    \[0\leq x\leq2\]
  7. Combining both domain conditions,
    \[\boxed{0\leq x\leq1}\]
  8. Step 2: Analyse the Range of the Terms
  9. For
    \[0\leq x\leq1,\]
    we have
    \[0\leq\sin^{-1}x\leq\dfrac{\pi}{2}\]
  10. Therefore,
    \[2\sin^{-1}x\geq0\]
  11. Also, since
    \[-1\leq1-x\leq1\]
  12. the principal value of the inverse sine satisfies
    \[-\dfrac{\pi}{2}\leq\sin^{-1}(1-x)\leq\dfrac{\pi}{2}\]
  13. Hence,
    \[\sin^{-1}(1-x)\leq\dfrac{\pi}{2}\]
  14. Step 3: Use the Given Equation
  15. The equation is
    \[\sin^{-1}(1-x)-2\sin^{-1}x=\dfrac{\pi}{2}\]
  16. Rearranging,
    \[\sin^{-1}(1-x)=\dfrac{\pi}{2}+2\sin^{-1}x\]
  17. But we already know that
    \[\sin^{-1}(1-x)\leq\dfrac{\pi}{2}\]
  18. Therefore, the right-hand side must also satisfy
    \[\dfrac{\pi}{2}+2\sin^{-1}x\leq\dfrac{\pi}{2}\]
  19. Subtracting \(\dfrac{\pi}{2}\) from both sides,
    \[2\sin^{-1}x\leq0\]
  20. But from \(x\geq0\), we have
    \[\sin^{-1}x\geq0\]
  21. Therefore, both inequalities can hold simultaneously only when
    \[\sin^{-1}x=0\]
  22. Hence,
    \[x=0\]
  23. Therefore, the correct option is
    \[\boxed{\text{C) }0}\]
🎯 Exam Significance
Exam Significance
›

This question is an excellent test of understanding the principal value range of inverse trigonometric functions. A lengthy trigonometric manipulation is unnecessary.

The decisive observation is

\[ \sin^{-1}(1-x)\leq\dfrac{\pi}{2}, \]

whereas the equation requires

\[ \sin^{-1}(1-x) = \dfrac{\pi}{2}+2\sin^{-1}x. \]

Since \(x\geq0\), the quantity \(2\sin^{-1}x\) cannot be negative. Therefore, equality is possible only when

\[ \sin^{-1}x=0. \]
This gives \(x=0\).

Significance for Competitive Entrance Examinations

The fastest approach is to use range analysis rather than applying trigonometric identities. From

\[ \sin^{-1}(1-x) = \dfrac{\pi}{2}+2\sin^{-1}x, \]

the right-hand side is at least \(\dfrac{\pi}{2}\). But the left-hand side can never exceed \(\dfrac{\pi}{2}\).

Therefore, equality is possible only at the boundary value

\[ \sin^{-1}x=0. \]

This range-based method is particularly useful in MCQs because it avoids unnecessary algebra and identifies the answer almost immediately.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. The common domain of the equation is \(0\leq x\leq1\).

  2. The principal value range of \(\sin^{-1}y\) is \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\).

  3. Hence, \(\sin^{-1}(1-x)\leq\dfrac{\pi}{2}\).

  4. For \(x\geq0\), \(\sin^{-1}x\geq0\).

  5. Thus, \(\dfrac{\pi}{2}+2\sin^{-1}x\) can equal at most \(\dfrac{\pi}{2}\) only when \(\sin^{-1}x=0\).

  6. Therefore, \(x=0\).

  7. Verification confirms that \(x=0\) satisfies the original equation.

  8. The correct option is \(\boxed{\text{C) }0}\).
← Q13
14 / 14  ·  100%
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Chapter Complete!

All 14 solutions for Inverse Trigonometric Functions covered.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 12 Maths Ch-2 Miscellaneous Solutions
NCERT Class 12 Maths Ch-2 Miscellaneous Solutions — Complete Notes & Solutions · academia-aeternum.com
Explore the NCERT Class 12 Mathematics Chapter 2 Miscellaneous Exercise: Inverse Trigonometric Functions with detailed, step-by-step solutions designed for both CBSE Board examinations and competitive entrance exams. This collection covers important problems based on principal values, domains and ranges, inverse trigonometric identities, half-angle and double-angle formulas, and transformations involving \(\sin^{-1}x\), \(\cos^{-1}x\), \(\tan^{-1}x\), and \(\cot^{-1}x\). Each solution explains…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    The Miscellaneous Exercise covers important problems based on inverse trigonometric functions, principal values, identities, equations, and transformations involving sin?¹x, cos?¹x, tan?¹x, and cot?¹x.

    Yes. Each solution is presented step by step with the required formulas, range conditions, domain restrictions, substitutions, simplification, and verification wherever necessary.

    Principal values determine the unique value returned by an inverse trigonometric function. They are essential when simplifying expressions such as sin?¹(sin x), cos?¹(cos x), and tan?¹(tan x).

    The exercise includes problems involving sin?¹x, cos?¹x, tan?¹x, and cot?¹x, along with their identities, principal values, domains, ranges, and related trigonometric equations.

    Yes. The solutions explain the concepts and intermediate steps required for CBSE Class 12 Mathematics, making them useful for revision, written-answer practice, and examination preparation.

    Yes. The problems develop skills in principal-value analysis, inverse-trigonometric identities, domain restrictions, and algebraic transformations that are useful for JEE and other competitive entrance examinations.

    First determine the domain, then use principal-value ranges, appropriate inverse-trigonometric identities, and standard trigonometric formulas. Always verify the final values in the original equation.

    Key concepts include principal values, domains and ranges, inverse-trigonometric identities, double-angle and half-angle formulas, trigonometric transformations, and verification of solutions.

    Domain and range checks prevent invalid substitutions and incorrect principal values. They are especially important when applying identities such as tan?¹(tan x), cos?¹(cos x), and cot?¹(cot x).

    They combine theory, a solution roadmap, detailed calculations, exam significance, and key takeaways, helping students understand the method as well as reach the correct answer efficiently.

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