Ch 2  ·  Q–
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Chapter 2 Exercise 2.1 Solutions

Inverse Trigonometric Functions

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 2 Exercise 2.1

Class 12 Mathematics Exercise 2.1 NCERT Solutions Inverse Trigonometric Functions Class 12 Mathematics Chapter 2 CBSE Board Exam JEE Main CUET Principal Values Inverse Trigonometric Functions Inverse Sine Inverse Cosine Inverse Tangent Inverse Secant Inverse Cosecant Inverse Cotangent
14 Questions
30–45 min Ideal time
Q1 Now at
Begin Q1 ↑ Progress bar
Q1
NUMERIC3 marks
Find the principal value of the following: \[\sin^{-1}\left(-\frac{1}{2}\right)\]
📘 Concept & Theory
Concept/Theory
›

The inverse trigonometric function \(\sin^{-1}x\), also written as \(\arcsin x\), gives the unique angle \(y\) whose sine is \(x\), subject to the principal value restriction.

For the inverse sine function, the principal value is always selected from the interval:

\[ -\frac{\pi}{2}\leq \sin^{-1}x\leq\frac{\pi}{2} \]

Therefore, the principal value range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).

We know that:

\[ \sin\frac{\pi}{6}=\frac{1}{2} \]

Since sine is an odd function,

\[ \sin(-\theta)=-\sin\theta \]

Hence,

\[ \sin\left(-\frac{\pi}{6}\right) = -\sin\frac{\pi}{6} = -\frac{1}{2} \]

Thus, \(-\frac{\pi}{6}\) is an angle whose sine is \(-\frac{1}{2}\). It also lies within the principal value range of inverse sine, so it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Convert the inverse trigonometric equation into an ordinary trigonometric equation.

  3. Identify the standard angle whose sine has magnitude \(\frac{1}{2}\).

  4. Use the negative sign to select the corresponding negative angle.

  5. Check that the obtained angle lies in the principal value range of \(\sin^{-1}x\).

  6. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of sin⁻¹(−1/2) y x −π/6 (√3/2, −1/2) sin(−π/6) = −1/2 Principal range of sin⁻¹x: [−π/2, π/2]
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Let
    \[ y=\sin^{-1}\left(-\frac{1}{2}\right)\]
  2. By the definition of the inverse sine function,
    \[ \sin y=-\frac{1}{2}\]
  3. We know the standard trigonometric value:
    \[\sin\frac{\pi}{6}=\frac{1}{2}\]
  4. Since sine is an odd function,
    \[\sin(-\theta)=-\sin\theta\]
  5. Therefore,
    \[\sin\left(-\frac{\pi}{6}\right)=-\sin\frac{\pi}{6}\]
    \[\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}\]
  6. Hence, an angle satisfying \(\sin y=-\frac{1}{2}\) is
    \[y=-\frac{\pi}{6}\]
  7. However, for an inverse trigonometric function, we must ensure that the angle obtained belongs to the principal value range.
  8. The principal value range of \(\sin^{-1}x\) is
    \[ -\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\]
  9. Since,
    \[-\frac{\pi}{2} < -\frac{\pi}{6} < \frac{\pi}{2}\]
  10. the value \(-\frac{\pi}{6}\) lies within the required principal value range.
  11. Therefore, the principal value is
    \[\boxed{\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}}\]
🎯 Exam Significance
Exam Significance
›

This question tests one of the most fundamental concepts of inverse trigonometric functions: the selection of the principal value. In board examinations, students are often expected to distinguish between all possible angles satisfying a trigonometric equation and the single principal value returned by an inverse trigonometric function.

A common mistake is to write an angle such as \(\frac{11\pi}{6}\) because it also has sine value \(-\frac{1}{2}\). However, \(\frac{11\pi}{6}\notin \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), so it cannot be the principal value of \(\sin^{-1}x\).

Remembering the principal value range of each inverse trigonometric function is therefore essential for solving direct-value questions, identities, equations and proof-based questions in the chapter.

Significance for Competitive Entrance Exams

For competitive examinations such as JEE and other entrance tests, principal value questions are frequently used as the foundation for more advanced problems involving inverse trigonometric identities, transformations and composite expressions.

The key skill is not merely recalling standard values but checking whether the angle belongs to the prescribed principal value branch. This becomes particularly important in expressions involving \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) and their combinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. \(\sin^{-1}x\) gives the principal angle whose sine is \(x\).

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

  3. \(\sin\frac{\pi}{6}=\frac{1}{2}\).

  4. Sine is an odd function, so \(\sin(-\theta)=-\sin\theta\).

  5. \(\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}\).

  6. The angle obtained must always be checked against the principal value range.

  7. Therefore, \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\).
↑ Top
1 / 14  ·  7%
Q2 →
Q2
NUMERIC3 marks
Find the principal value of the following:\[\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\]
📘 Concept & Theory
Concept/Theory
›

The inverse cosine function, \(\cos^{-1}x\), gives the unique principal value \(y\) such that

\[ \cos y=x \]

To make the cosine function one-one and therefore invertible, its domain is restricted to the interval

\[ [0,\pi] \]

Hence, the principal value range of the inverse cosine function is

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

We use the standard trigonometric value

\[ \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \]

Since \(\frac{\pi}{6}\) lies in the principal value range \([0,\pi]\), it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Use the definition of the inverse cosine function to write an equivalent cosine equation.

  3. Identify the standard angle whose cosine is \(\frac{\sqrt{3}}{2}\).

  4. Check that the angle lies within the principal value range of \(\cos^{-1}x\).

  5. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of cos⁻¹(√3/2) y x 1 −1 1 −1 π/6 (√3/2, 1/2) cos(π/6) = √3/2 Principal range of cos⁻¹x: [0, π]
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. Let
    \[y=\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\]
  2. By the definition of the inverse cosine function,
    \[\cos y=\frac{\sqrt{3}}{2}\]
  3. We know the standard trigonometric value
    \[\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\]
  4. Therefore, an angle satisfying \(\cos y=\frac{\sqrt{3}}{2}\) is
    \[y=\frac{\pi}{6}\]
  5. Now, for an inverse trigonometric function, the obtained angle must belong to its principal value range.
  6. The principal value range of the inverse cosine function is
    \[0\leq y\leq\pi\]
  7. Since
    \[0<\frac{\pi}{6}<\pi\]
  8. \(\frac{\pi}{6}\) lies within the principal value range of
    \[\cos^{-1}x\]
  9. Therefore, the principal value is
    \[\boxed{\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the fundamental relationship between a trigonometric function and its inverse function. For board examinations, it is important to remember that \(\cos^{-1}x\) does not give every angle having cosine equal to \(x\). It gives only the unique angle selected from its principal value range.

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

\[ [0,\pi] \]

Therefore, even if other angles have the same cosine value, they cannot be accepted as the principal value if they lie outside this interval. This concept is frequently used in direct-value questions as well as inverse trigonometric identities.

Significance for Competitive Entrance Exams

In competitive entrance examinations, principal value restrictions are particularly important when solving expressions involving multiple inverse trigonometric functions. A correct standard angle is not enough; its location with respect to the principal value interval must also be verified.

For this problem, the standard value \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\) immediately suggests \(\frac{\pi}{6}\). Since this angle belongs to \([0,\pi]\), no further adjustment is required.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. \(\cos^{-1}x\) returns the principal angle \(y\) satisfying \(\cos y=x\).

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

  3. \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\).

  4. \(\frac{\pi}{6}\) lies within the principal value range \([0,\pi]\).

  5. Therefore, no change or adjustment of the standard angle is required.

  6. Always check the principal value range when evaluating an inverse trigonometric function.
← Q1
2 / 14  ·  14%
Q3 →
Q3
NUMERIC3 marks
Find the principal value of the following: \[\operatorname{cosec}^{-1}(2)\]
📘 Concept & Theory
Concept/Theory
›

The inverse cosecant function, \(\operatorname{cosec}^{-1}x\), gives the principal value \(y\) for which

\[ \operatorname{cosec}y=x \]

Since

\[ \operatorname{cosec}y=\frac{1}{\sin y} \]

the equation can be converted into an equation involving the sine function.

For the principal value branch of the inverse cosecant function, the range is

\[ \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right] \]

The point \(0\) is excluded because \(\operatorname{cosec}0\) is undefined.

We know that

\[ \sin\frac{\pi}{6}=\frac{1}{2} \]

Therefore,

\[ \operatorname{cosec}\frac{\pi}{6} = \frac{1}{\sin\frac{\pi}{6}} = \frac{1}{\frac{1}{2}} = 2 \]

Since \(\frac{\pi}{6}\) belongs to the principal value range of \(\operatorname{cosec}^{-1}x\), it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Use the definition of the inverse cosecant function.

  3. Rewrite cosecant in terms of sine.

  4. Obtain the corresponding sine equation.

  5. Identify the standard angle satisfying the sine equation.

  6. Check that the angle lies in the principal value range of \(\operatorname{cosec}^{-1}x\).

  7. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of cosec⁻¹(2) y x 1 −1 1 −1 π/6 (√3/2, 1/2) cosec(π/6) = 2 Principal range of cosec⁻¹x: [−π/2, 0) ∪ (0, π/2]
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  12 steps
  1. Let
    \[y=\operatorname{cosec}^{-1}(2)\]
  2. By the definition of the inverse cosecant function,
    \[\operatorname{cosec}y=2\]
  3. We know that
    \[\operatorname{cosec}y=\frac{1}{\sin y}\]
  4. Therefore,
    \[\frac{1}{\sin y}=2\]
  5. Taking the reciprocal on both sides gives
    \[\sin y=\frac{1}{2}\]
  6. We know the standard trigonometric value
    \[\sin\frac{\pi}{6}=\frac{1}{2}\]
  7. Hence,
    \[y=\frac{\pi}{6}\]
  8. Now we must verify that this angle belongs to the principal value range of the inverse cosecant function.
  9. The principal value range of \(\operatorname{cosec}^{-1}x\) is
    \[\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]\]
  10. Since
    \[0<\frac{\pi}{6}\leq\frac{\pi}{2}\]
  11. we have
    \[\frac{\pi}{6}\in \left(0,\frac{\pi}{2}\right]\]
  12. Therefore, \(\frac{\pi}{6}\) is within the principal value range of \(\operatorname{cosec}^{-1}x\)
  13. Hence, the principal value is
    \[\boxed{\operatorname{cosec}^{-1}(2)=\frac{\pi}{6}}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it introduces the principal value branch of the inverse cosecant function. Unlike \(\sin^{-1}x\) and \(\cos^{-1}x\), the principal value range of \(\operatorname{cosec}^{-1}x\) contains two intervals.

A particularly important point is that \(0\) must be excluded from the range because

\[ \operatorname{cosec}0=\frac{1}{\sin0} \]

is undefined.

Therefore, writing the principal range simply as \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) is not strictly correct for the inverse cosecant function. The correct principal value range is

\[ \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right] \]

This distinction can be important in board examination questions involving inverse trigonometric functions and their properties.

Significance for Competitive Entrance Exams

For competitive entrance examinations, inverse cosecant questions often test whether the candidate remembers the correct principal value branch rather than simply recalling a standard trigonometric value.

The safest method is to first convert the inverse cosecant expression into an ordinary trigonometric equation and then verify the resulting angle against the principal value range.

For this problem,

\[ \operatorname{cosec}y=2 \]
\[ \frac{1}{\sin y}=2 \]
\[ \sin y=\frac{1}{2} \]
\[ y=\frac{\pi}{6} \]

and the final range check confirms that the answer is valid.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. \(\operatorname{cosec}^{-1}x\) gives the principal value \(y\) satisfying \(\operatorname{cosec}y=x\).

  2. \(\operatorname{cosec}y=\frac{1}{\sin y}\).

  3. For \(\operatorname{cosec}^{-1}x\), the principal value range is \(\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]\).

  4. The value \(0\) is excluded because \(\operatorname{cosec}0\) is undefined.

  5. \(\sin\frac{\pi}{6}=\frac{1}{2}\), so \(\operatorname{cosec}\frac{\pi}{6}=2\).

  6. \(\frac{\pi}{6}\) lies in the principal value range of \(\operatorname{cosec}^{-1}x\).

  7. Therefore, \(\operatorname{cosec}^{-1}(2)=\frac{\pi}{6}\).
← Q2
3 / 14  ·  21%
Q4 →
Q4
NUMERIC3 marks
Find the principal value of the following: \[\tan^{-1}\left(-\sqrt{3}\right)\]
📘 Concept & Theory
Concept/Theory
›

The inverse tangent function, \(\tan^{-1}x\), gives the unique principal value \(y\) satisfying

\[ \tan y=x \]

Since the tangent function is periodic and therefore not one-one over its entire domain, its domain is restricted to

\[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]

On this interval, the tangent function is one-one and assumes every real value. Hence, the principal value range of \(\tan^{-1}x\) is

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

We know the standard trigonometric value

\[ \tan\frac{\pi}{3}=\sqrt{3} \]

Since tangent is an odd function,

\[ \tan(-\theta)=-\tan\theta \]

Therefore,

\[ \tan\left(-\frac{\pi}{3}\right) = -\tan\frac{\pi}{3} = -\sqrt{3} \]

Since \(-\frac{\pi}{3}\) lies within the principal value range of \(\tan^{-1}x\), it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Use the definition of the inverse tangent function.

  3. Identify the standard angle whose tangent has magnitude \(\sqrt{3}\).

  4. Use the negative sign to obtain the corresponding negative angle.

  5. Check that the angle lies in the principal value range of \(\tan^{-1}x\).

  6. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of tan⁻¹(−√3) y x 1 −1 1 −1 −π/3 (1/2, −√3/2) tan(−π/3) = −√3 Principal range of tan⁻¹x: (−π/2, π/2)
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Let
    \[ y=\tan^{-1}\left(-\sqrt{3}\right)\]
  2. By the definition of the inverse tangent function,
  3. \[\tan y=-\sqrt{3}\]
  4. We know that
    \[\tan\frac{\pi}{3}=\sqrt{3}\]
  5. Since tangent is an odd function,
    \[\tan(-\theta)=-\tan\theta\]
  6. Therefore,
    \[\tan\left(-\frac{\pi}{3}\right)=-\tan\frac{\pi}{3}\]
    \[\tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}\]
  7. Hence, an angle satisfying \(\tan y=-\sqrt{3}\) is
    \[y=-\frac{\pi}{3}\]
  8. We must now verify that this angle belongs to the principal value range of the inverse tangent function.
  9. The principal value range of \(\tan^{-1}x\) is
    \[-\frac{\pi}{2} < y < \frac{\pi}{2}\]
  10. Since
    \[ -\frac{\pi}{2} < -\frac{\pi}{3} < \frac{\pi}{2}\]
    the angle \(-\frac{\pi}{3}\) lies within the principal value range.
  11. Therefore, the principal value is
    \[\boxed{\tan^{-1}\left(-\sqrt{3}\right)=-\frac{\pi}{3}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the standard values of the tangent function together with the principal value concept of inverse trigonometric functions. For board examinations, it is important to remember that \(\tan^{-1}x\) returns only one angle from its principal value range.

The principal value range is

\[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]

Thus, although infinitely many angles can have tangent equal to \(-\sqrt{3}\), only the angle lying in this interval can be the principal value of \(\tan^{-1}(-\sqrt{3})\).

Students should also remember that the endpoints \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) are excluded because tangent is undefined at these angles.

Significance for Competitive Entrance Exams

In competitive entrance examinations, principal value selection is frequently combined with identities involving inverse trigonometric functions. A common source of errors is identifying a correct trigonometric angle without checking whether it belongs to the prescribed principal value interval.

Here, the negative value immediately indicates that the principal angle must lie in the negative part of the tangent principal range. The standard value

\[ \tan\frac{\pi}{3}=\sqrt{3} \]

therefore leads directly to

\[ \tan\left(-\frac{\pi}{3}\right)=-\sqrt{3} \]

and the range check confirms the result.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(\tan^{-1}x\) gives the principal angle \(y\) satisfying \(\tan y=x\).

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

  3. \(\tan\frac{\pi}{3}=\sqrt{3}\).

  4. Tangent is an odd function, so \(\tan(-\theta)=-\tan\theta\).

  5. \(\tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}\).

  6. \(-\frac{\pi}{3}\) lies within the principal value range of \(\tan^{-1}x\).

  7. The endpoints \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) are excluded because tangent is undefined there.

  8. Therefore, \(\tan^{-1}\left(-\sqrt{3}\right)=-\frac{\pi}{3}\).
← Q3
4 / 14  ·  29%
Q5 →
Q5
NUMERIC3 marks
Find the principal value of the following: \[\cos^{-1}\left(-\frac{1}{2}\right)\]
📘 Concept & Theory
Concept/Theory
›

The inverse cosine function, \(\cos^{-1}x\), gives the unique principal value \(y\) satisfying

\[ \cos y=x \]

Since the cosine function is not one-one over its complete domain \(\mathbb{R}\), its domain is restricted to the interval

\[ [0,\pi] \]

On this interval, cosine is one-one and takes every value from \(1\) to \(-1\). Therefore, the principal value range of \(\cos^{-1}x\) is

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

We know the standard value

\[ \cos\frac{\pi}{3}=\frac{1}{2} \]

Using the supplementary-angle identity

\[ \cos(\pi-\theta)=-\cos\theta \]

we obtain

\[ \cos\left(\pi-\frac{\pi}{3}\right) = -\cos\frac{\pi}{3} \]
\[ \cos\frac{2\pi}{3} = -\frac{1}{2} \]

Since \(\frac{2\pi}{3}\) lies in the principal value interval \([0,\pi]\), it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Convert the inverse cosine equation into an ordinary cosine equation.

  3. Identify the standard angle having cosine \(\frac{1}{2}\).

  4. Use the supplementary-angle identity to obtain an angle whose cosine is \(-\frac{1}{2}\).

  5. Verify that the resulting angle belongs to the principal value range of \(\cos^{-1}x\).

  6. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of cos⁻¹(−1/2) 2π/3 x = −1/2 y x Principal range of cos⁻¹x: [0, π]
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Let
    \[y=\cos^{-1}\left(-\frac{1}{2}\right)\]
  2. By the definition of the inverse cosine function,
    \[\cos y=-\frac{1}{2}\]
  3. We know that
    \[\cos\frac{\pi}{3}=\frac{1}{2}\]
  4. The required cosine value is negative. Using the identity
    \[\cos(\pi-\theta)=-\cos\theta\]
  5. put \(\theta=\frac{\pi}{3}\). Then
    \[\cos\left(\pi-\frac{\pi}{3}\right)=-\cos\frac{\pi}{3}\]
  6. Simplifying the angle,
    \[\begin{aligned}\pi-\frac{\pi}{3}&=\frac{3\pi-\pi}{3}\\&=\frac{2\pi}{3}\end{aligned}\]
  7. Therefore,
    \[\cos\frac{2\pi}{3}=-\frac{1}{2}\]
  8. Hence, an angle satisfying \(\cos y=-\frac{1}{2}\) is
    \[y=\frac{2\pi}{3}\]
  9. We now verify the principal value condition.
  10. The principal value range of \(\cos^{-1}x\) is
    \[0\leq y\leq\pi\]
  11. Since
    \[0<\frac{2\pi}{3}<\pi\]
    the angle \(\frac{2\pi}{3}\) lies within the principal value range.
  12. Therefore, the principal value is
    \[\boxed{\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it demonstrates how the sign of a trigonometric value determines the location of the principal angle. For \(\cos^{-1}x\), the principal value must always lie in \([0,\pi]\).

Since cosine is negative in the second quadrant, the principal angle corresponding to \(-\frac{1}{2}\) must lie between \(\frac{\pi}{2}\) and \(\pi\). Therefore, the standard angle \(\frac{\pi}{3}\) cannot be the answer because

\[ \cos\frac{\pi}{3}=\frac{1}{2} \]

The corresponding second-quadrant angle is

\[ \pi-\frac{\pi}{3}=\frac{2\pi}{3} \]

This quadrant-based reasoning is particularly useful in board questions involving direct evaluation and inverse trigonometric identities.

Significance for Competitive Entrance Exams

Competitive examinations often test inverse trigonometric functions by combining standard values, quadrant information and principal value restrictions. A student must distinguish between all possible solutions of a trigonometric equation and the single value returned by the inverse function.

Here, although infinitely many angles can have cosine equal to \(-\frac{1}{2}\), the principal value must belong to \([0,\pi]\). Thus,

\[ \cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3} \]

A useful mental check is that cosine is negative in the second quadrant, and \(\frac{2\pi}{3}\) lies in that quadrant.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(\cos^{-1}x\) gives the principal angle \(y\) satisfying \(\cos y=x\).

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

  3. \(\cos\frac{\pi}{3}=\frac{1}{2}\).

  4. Cosine is negative in the second quadrant.

  5. \(\cos(\pi-\theta)=-\cos\theta\).

  6. \(\cos\frac{2\pi}{3}=-\frac{1}{2}\).

  7. \(\frac{2\pi}{3}\) belongs to the principal value range of \(\cos^{-1}x\).

  8. Therefore, \(\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}\).
← Q4
5 / 14  ·  36%
Q6 →
Q6
NUMERIC3 marks
Find the principal value of the following: \[\tan^{-1}(-1)\]
📘 Concept & Theory
Concept/Theory
›

The inverse tangent function, \(\tan^{-1}x\), gives the unique principal value \(y\) satisfying

\[ \tan y=x \]

The tangent function is periodic and is therefore not one-one over its complete domain. To define its inverse, its domain is restricted to

\[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]

On this interval, \(\tan x\) is one-one and takes every real value. Hence, the principal value range of \(\tan^{-1}x\) is

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

We know the standard value

\[ \tan\frac{\pi}{4}=1 \]

Since tangent is an odd function,

\[ \tan(-\theta)=-\tan\theta \]

Therefore,

\[ \tan\left(-\frac{\pi}{4}\right) = -\tan\frac{\pi}{4} = -1 \]

Thus, \(-\frac{\pi}{4}\) is the angle whose tangent is \(-1\), and it lies within the principal value range.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Convert the inverse tangent equation into an ordinary tangent equation.

  3. Recall the standard value \(\tan\frac{\pi}{4}=1\).

  4. Use the odd-function property of tangent to obtain a negative value.

  5. Check that the resulting angle lies in the principal value range of \(\tan^{-1}x\).

  6. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of tan⁻¹(−1) y x 1 −1 1 −1 −π/4 (1/√2, −1/√2) tan(−π/4) = −1 Principal range of tan⁻¹x: (−π/2, π/2)
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Let
    \[y=\tan^{-1}(-1)\]
  2. By the definition of the inverse tangent function,
    \[\tan y=-1\]
  3. We know that
    \[\tan\frac{\pi}{4}=1\]
  4. Since tangent is an odd function,
    \[\tan(-\theta)=-\tan\theta\]
  5. Therefore,
    \[\tan\left(-\frac{\pi}{4}\right)=-\tan\frac{\pi}{4}\]
    \[\tan\left(-\frac{\pi}{4}\right)=-1\]
  6. Hence, an angle satisfying \(\tan y=-1\) is
    \[y=-\frac{\pi}{4}\]
  7. We now verify the principal value condition.
  8. The principal value range of \(\tan^{-1}x\) is
    \[-\frac{\pi}{2} < y < \frac{\pi}{2}\]
  9. Since
    \[-\frac{\pi}{2} < -\frac{\pi}{4} < \frac{\pi}{2}\]
    the angle \(-\frac{\pi}{4}\) lies within the principal value range.
  10. Therefore, the principal value is
    \[\boxed{\tan^{-1}(-1)=-\frac{\pi}{4}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the standard value of tangent and the principal value range of its inverse function. A common error is to confuse the principal value range of tangent with that of another trigonometric function.

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

\[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]

It is important to note that the endpoints are excluded because \(\tan x\) is undefined at \(\frac{\pi}{2}\) and \(-\frac{\pi}{2}\).

Also, the angle must be written in radians when giving the final answer in standard NCERT form. Thus, although \(-45^\circ\) is equivalent to \(-\frac{\pi}{4}\), the preferred answer is

\[ -\frac{\pi}{4} \]
Significance for Competitive Entrance Exams

This basic result is frequently used as a building block in questions involving inverse trigonometric identities and expressions such as \(\tan^{-1}x+\tan^{-1}y\).

Competitive examination problems may also deliberately include different-looking angles that have the same tangent. The principal value restriction determines which one is the correct value of the inverse function.

Here, although

\[ \tan\left(-\frac{\pi}{4}+n\pi\right)=-1, \qquad n\in\mathbb{Z}, \]

only

\[ -\frac{\pi}{4} \]

belongs to the principal value range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(\tan^{-1}x\) gives the principal angle \(y\) satisfying \(\tan y=x\).

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

  3. \(\tan\frac{\pi}{4}=1\).

  4. Tangent is an odd function, so \(\tan(-\theta)=-\tan\theta\).

  5. \(\tan\left(-\frac{\pi}{4}\right)=-1\).

  6. \(-\frac{\pi}{4}\) lies within the principal value range of \(\tan^{-1}x\).

  7. The endpoints of the principal range are excluded.

  8. Therefore, \(\tan^{-1}(-1)=-\frac{\pi}{4}\).
← Q5
6 / 14  ·  43%
Q7 →
Q7
NUMERIC3 marks
Find the principal value of the following: \[\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\]
📘 Concept & Theory
Concept/Theory
›

The inverse secant function, \(\sec^{-1}x\), gives the principal value \(y\) satisfying

\[ \sec y=x \]

Since

\[ \sec y=\frac{1}{\cos y} \]

the equation involving secant can be converted into an equation involving cosine.

The principal value range of the inverse secant function is

\[ [0,\pi]\setminus\left\{\frac{\pi}{2}\right\} \]

Equivalently, it can be written as

\[ \left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right] \]

The value \(\frac{\pi}{2}\) is excluded because \(\sec\frac{\pi}{2}\) is undefined.

We know the standard value

\[ \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \]

Therefore,

\[ \sec\frac{\pi}{6} = \frac{1}{\cos\frac{\pi}{6}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \]

Hence, \(\frac{\pi}{6}\) is an angle whose secant is \(\frac{2}{\sqrt{3}}\). Since it lies in the principal value range of \(\sec^{-1}x\), it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Use the definition of the inverse secant function.

  3. Rewrite secant as the reciprocal of cosine.

  4. Solve the resulting cosine equation using a standard trigonometric value.

  5. Check that the obtained angle belongs to the principal value range of \(\sec^{-1}x\).

  6. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of sec⁻¹(2/√3) y x 1 −1 1 −1 π/6 (√3/2, 1/2) sec(π/6) = 2/√3 Principal range of sec⁻¹x: [0, π] \ {π/2}
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  12 steps
  1. Let
    \[y=\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\]
  2. By the definition of the inverse secant function,
    \[\sec y=\frac{2}{\sqrt{3}}\]
  3. Since
    \[\sec y=\frac{1}{\cos y}\]
  4. we have
    \[\frac{1}{\cos y}=\frac{2}{\sqrt{3}}\]
  5. Taking the reciprocal of both sides,
    \[\cos y=\frac{\sqrt{3}}{2}\]
  6. We know the standard trigonometric value
    \[\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\]
  7. Therefore,
    \[y=\frac{\pi}{6}\]
  8. We now verify that this value satisfies the principal value condition.
  9. The principal value range of \(\sec^{-1}x\) is
    \[[0,\pi]\setminus\left\{\frac{\pi}{2}\right\}\]
  10. Since
    \[0<\frac{\pi}{6}<\frac{\pi}{2}\]
  11. we have
    \[\frac{\pi}{6}\in \left[0,\frac{\pi}{2}\right)\]
  12. Hence, \(\frac{\pi}{6}\) belongs to the principal value range of \(\sec^{-1}x\)
  13. Therefore, the principal value is
    \[\boxed{\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the relationship between secant and cosine and the principal value range of the inverse secant function. The important transformation is

\[ \sec y=\frac{1}{\cos y} \]

which allows the problem to be reduced to a standard cosine value.

Students should be careful with the principal value range. It is not simply the entire interval \([0,\pi]\), because \(\sec\frac{\pi}{2}\) is undefined. The correct range is

\[ [0,\pi]\setminus\left\{\frac{\pi}{2}\right\} \]

This distinction is useful in board questions involving inverse secant, inverse cosecant and their identities.

Significance for Competitive Entrance Exams

In competitive entrance examinations, inverse secant expressions may appear in combinations with inverse cosine or other inverse trigonometric functions. Converting secant into the reciprocal of cosine often provides the quickest route to the answer.

Here,

\[ \sec y=\frac{2}{\sqrt{3}} \]

immediately gives

\[ \cos y=\frac{\sqrt{3}}{2} \]

and hence

\[ y=\frac{\pi}{6} \]

The final range check confirms that this is the principal value.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(\sec^{-1}x\) gives the principal angle \(y\) satisfying \(\sec y=x\).

  2. \(\sec y=\frac{1}{\cos y}\).

  3. The principal value range of \(\sec^{-1}x\) is \([0,\pi]\setminus\left\{\frac{\pi}{2}\right\}\).

  4. \(\frac{\pi}{2}\) is excluded because \(\sec\frac{\pi}{2}\) is undefined.

  5. \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\).

  6. Therefore, \(\sec\frac{\pi}{6}=\frac{2}{\sqrt{3}}\).

  7. \(\frac{\pi}{6}\) lies within the principal value range of \(\sec^{-1}x\).

  8. Therefore, \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}\).
← Q6
7 / 14  ·  50%
Q8 →
Q8
NUMERIC3 marks
Find the principal value of the following: \[\cot^{-1}\left(\sqrt{3}\right)\]
📘 Concept & Theory
Concept/Theory
›

The inverse cotangent function, \(\cot^{-1}x\), gives the unique principal value \(y\) satisfying

\[ \cot y=x \]

For the principal value branch used in NCERT, the range of \(\cot^{-1}x\) is

\[ (0,\pi) \]

The endpoints \(0\) and \(\pi\) are excluded because cotangent is undefined at both these angles.

We know the standard trigonometric value

\[ \tan\frac{\pi}{6}=\frac{1}{\sqrt{3}} \]

Since cotangent is the reciprocal of tangent,

\[ \cot\theta=\frac{1}{\tan\theta} \]

therefore,

\[ \cot\frac{\pi}{6} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \]

Since \(\frac{\pi}{6}\) lies in the principal value range \((0,\pi)\), it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Use the definition of the inverse cotangent function.

  3. Recall the standard value of \(\tan\frac{\pi}{6}\).

  4. Use the reciprocal relationship between cotangent and tangent.

  5. Verify that the obtained angle belongs to the principal value range of \(\cot^{-1}x\).

  6. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of cot⁻¹(√3) y x 1 −1 1 −1 π/6 (√3/2, 1/2) cot(π/6) = √3 Principal range of cot⁻¹x: (0, π)
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. Let
    \[y=\cot^{-1}\left(\sqrt{3}\right)\]
  2. By the definition of the inverse cotangent function,
    \[\cot y=\sqrt{3}\]
  3. We know that cotangent is the reciprocal of tangent:
    \[\cot y=\frac{1}{\tan y}\]
  4. We also know the standard trigonometric value
    \[\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\]
  5. Therefore,
    \[\begin{aligned}\cot\frac{\pi}{6}&=\frac{1}{\tan\frac{\pi}{6}}\cot\frac{\pi}{6}\\&=\frac{1}{\frac{1}{\sqrt{3}}}\\& =\cot\frac{\pi}{6}=\sqrt{3}\end{aligned}\]
  6. Hence, an angle satisfying \(\cot y=\sqrt{3}\) is
    \[y=\frac{\pi}{6}\]
  7. We now verify the principal value condition.
  8. The principal value range of \(\cot^{-1}x\) is
    \[0 < y < \pi\]
  9. Since
    \[0<\frac{\pi}{6}<\pi\]
  10. the angle \(\frac{\pi}{6}\) lies within the principal value range.
  11. Therefore, the principal value is
    \[\boxed{\cot^{-1}\left(\sqrt{3}\right)=\frac{\pi}{6}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the standard values of trigonometric functions and the principal value range of the inverse cotangent function. A useful relationship to remember is

\[ \cot\theta=\frac{1}{\tan\theta} \]

Thus, when a cotangent value is given, it can often be evaluated quickly by recalling a corresponding tangent value.

Students should also remember that the principal value range of \(\cot^{-1}x\) is

\[ (0,\pi) \]

The endpoints are excluded because \(\cot0\) and \(\cot\pi\) are undefined. This range restriction is essential when identifying the principal value.

Significance for Competitive Entrance Exams

For competitive entrance examinations, inverse cotangent values are often combined with inverse tangent and other inverse trigonometric functions. Recognising reciprocal relationships can substantially reduce the calculation time.

In this problem,

\[ \cot\frac{\pi}{6}=\sqrt{3} \]

immediately identifies the required angle. The range check then confirms that \(\frac{\pi}{6}\) is the principal value.

It is important not to treat \(\cot^{-1}x\) as simply the reciprocal of \(\tan^{-1}x\). The notation \(\cot^{-1}x\) denotes the inverse function, not the reciprocal of the inverse tangent.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(\cot^{-1}x\) gives the principal angle \(y\) satisfying \(\cot y=x\).

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

  3. \(\cot\theta=\frac{1}{\tan\theta}\).

  4. \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\).

  5. Therefore, \(\cot\frac{\pi}{6}=\sqrt{3}\).

  6. \(\frac{\pi}{6}\) lies within the principal value range of \(\cot^{-1}x\).

  7. The notation \(\cot^{-1}x\) represents the inverse cotangent function, not the reciprocal of \(\tan^{-1}x\).

  8. Therefore, \(\cot^{-1}\left(\sqrt{3}\right)=\frac{\pi}{6}\).
← Q7
8 / 14  ·  57%
Q9 →
Q9
NUMERIC3 marks
Find the principal value of the following: \[\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\]
📘 Concept & Theory
Concept/Theory
›

The inverse cosine function, \(\cos^{-1}x\), gives the unique principal value \(y\) satisfying

\[ \cos y=x \]

The principal value range of the inverse cosine function is

\[ 0\leq y\leq\pi \]

Since

\[ \cos\frac{\pi}{4}=\frac{1}{\sqrt{2}} \]

and cosine is negative in the second quadrant, the corresponding principal angle is obtained using

\[ \pi-\frac{\pi}{4}=\frac{3\pi}{4} \]

Therefore,

\[ \cos\frac{3\pi}{4} = -\frac{1}{\sqrt{2}} \]

Since \(\frac{3\pi}{4}\in[0,\pi]\), it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Convert the inverse cosine expression into an ordinary cosine equation.

  3. Recall the standard value \(\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}\).

  4. Since the required cosine value is negative, select the corresponding angle in the second quadrant.

  5. Use the supplementary-angle relation to obtain \(\frac{3\pi}{4}\).

  6. Verify that the angle lies in the principal value range of \(\cos^{-1}x\).

  7. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of cos⁻¹(−1/√2) y x 1 −1 1 −1 3π/4 (−1/√2, 1/√2) cos(3π/4) = −1/√2 Principal range of cos⁻¹x: [0, π]
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  12 steps
  1. Let
    \[y=\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\]
  2. By the definition of the inverse cosine function,
    \[\cos y=-\frac{1}{\sqrt{2}}\]
  3. We know the standard trigonometric value
    \[\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}\]
  4. The required value of cosine is negative. In the principal value range \([0,\pi]\), cosine is negative in the second quadrant.
  5. The second-quadrant angle corresponding to the reference angle \(\frac{\pi}{4}\) is
    \[\pi-\frac{\pi}{4}\]
  6. Simplifying,
    \[\begin{aligned}\pi-\frac{\pi}{4}&=\frac{4\pi-\pi}{4}\\&=\frac{3\pi}{4}\end{aligned}\]
  7. Therefore,
    \[\cos\frac{3\pi}{4}=\cos\left(\pi-\frac{\pi}{4}\right)\]
  8. Using
    \[\cos(\pi-\theta)=-\cos\theta\]
  9. we get
    \[\begin{aligned}\cos\frac{3\pi}{4}&=-\cos\frac{\pi}{4}\cos\frac{3\pi}{4}\\&=-\frac{1}{\sqrt{2}}\end{aligned} \]
  10. Hence, an angle satisfying \(\cos y=-\frac{1}{\sqrt{2}}\) is
    \[y=\frac{3\pi}{4}\]
  11. We now verify the principal value condition.
  12. The principal value range of \(\cos^{-1}x\) is
    \[0\leq y\leq\pi\]
  13. Since
    \[0<\frac{3\pi}{4}<\pi\]
    the angle \(\frac{3\pi}{4}\) lies within the principal value range.
  14. Therefore, the principal value is
    \[\boxed{\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi}{4}}\]
🎯 Exam Significance
Exam Significance
›

This question reinforces two important concepts: standard trigonometric values and principal value selection. Because the argument of \(\cos^{-1}\) is negative, the principal angle must lie in the second quadrant.

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

\[ [0,\pi] \]

Within this interval, cosine is positive in the first quadrant and negative in the second quadrant. Therefore, the reference angle \(\frac{\pi}{4}\) must be converted to its corresponding second-quadrant angle:

\[ \pi-\frac{\pi}{4}=\frac{3\pi}{4} \]

This quadrant-based approach is especially useful for direct-value questions in board examinations.

Significance for Competitive Entrance Exams

In competitive entrance examinations, the ability to determine the correct quadrant quickly is essential when evaluating inverse trigonometric expressions. Instead of searching through multiple possible angles, students can identify the reference angle and then select the angle permitted by the principal value range.

Here, the reference angle is

\[ \frac{\pi}{4} \]

and the negative cosine requires the second quadrant. Hence,

\[ y=\pi-\frac{\pi}{4}=\frac{3\pi}{4} \]

The range check confirms that this is the principal value.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(\cos^{-1}x\) gives the principal angle \(y\) satisfying \(\cos y=x\).

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

  3. \(\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}\).

  4. Cosine is negative in the second quadrant.

  5. The second-quadrant angle corresponding to reference angle \(\frac{\pi}{4}\) is \(\pi-\frac{\pi}{4}=\frac{3\pi}{4}\).

  6. \(\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}\).

  7. \(\frac{3\pi}{4}\) lies within the principal value range of \(\cos^{-1}x\).

  8. Therefore, \(\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi}{4}\).
← Q8
9 / 14  ·  64%
Q10 →
Q10
NUMERIC3 marks
Find the principal value of the following: \[\operatorname{cosec}^{-1}\left(-\sqrt{2}\right)\]
📘 Concept & Theory
Concept/Theory
›

The inverse cosecant function, \(\operatorname{cosec}^{-1}x\), gives the principal value \(y\) satisfying

\[ \operatorname{cosec}y=x \]

Since

\[ \operatorname{cosec}y=\frac{1}{\sin y} \]

an inverse cosecant equation can be converted into an equation involving the sine function.

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

\[ \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right] \]

The value \(0\) is excluded because \(\operatorname{cosec}0\) is undefined.

We know that

\[ \sin\frac{\pi}{4}=\frac{1}{\sqrt{2}} \]

Since sine is an odd function,

\[ \sin(-\theta)=-\sin\theta \]

therefore,

\[ \sin\left(-\frac{\pi}{4}\right) = -\sin\frac{\pi}{4} = -\frac{1}{\sqrt{2}} \]

Consequently,

\[ \operatorname{cosec}\left(-\frac{\pi}{4}\right) = \frac{1}{-\frac{1}{\sqrt{2}}} = -\sqrt{2} \]

Since \(-\frac{\pi}{4}\) belongs to the principal value range of \(\operatorname{cosec}^{-1}x\), it is the required principal value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the required principal value be \(y\).

  2. Use the definition of the inverse cosecant function.

  3. Rewrite cosecant in terms of sine.

  4. Obtain the corresponding sine equation.

  5. Use the standard value of \(\sin\frac{\pi}{4}\).

  6. Select the negative angle because the required sine value is negative.

  7. Check that the angle lies in the principal value range of \(\operatorname{cosec}^{-1}x\).

  8. State the principal value.

📊 Graph / Figure
Graph / Figure
›
Principal Value of cosec⁻¹(−√2) y x 1 −1 1 −1 −π/4 (1/√2, −1/√2) cosec(−π/4) = −√2 PV range: [−π/2, 0) ∪ (0, π/2]
Relevant Visual Representation
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Let
    \[y=\operatorname{cosec}^{-1}\left(-\sqrt{2}\right)\]
  2. By the definition of the inverse cosecant function,
    \[\operatorname{cosec}y=-\sqrt{2}\]
  3. Since
    \[\operatorname{cosec}y=\frac{1}{\sin y}\]
  4. we have
    \[\frac{1}{\sin y}=-\sqrt{2}\]
  5. Taking the reciprocal of both sides,
    \[\sin y=-\frac{1}{\sqrt{2}}\]
  6. We know the standard trigonometric value
    \[\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\]
  7. Since the required sine value is negative, we consider the corresponding negative angle:
    \[\begin{aligned}\sin\left(-\frac{\pi}{4}\right)&=-\sin\frac{\pi}{4}\sin\left(-\frac{\pi}{4}\right)\\&=-\frac{1}{\sqrt{2}}\end{aligned}\]
  8. Therefore,
    \[y=-\frac{\pi}{4}\]
  9. We now verify that this value belongs to the principal value range of \(\operatorname{cosec}^{-1}x\)
  10. The principal value range is
    \[\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]\]
  11. Since
    \[-\frac{\pi}{2} < -\frac{\pi}{4} < 0\]
  12. we have
    \[-\frac{\pi}{4}\in \left[-\frac{\pi}{2},0\right)\]
  13. Hence, \(-\frac{\pi}{4}\) is within the principal value range.
  14. Therefore, the principal value is
    \[\boxed{\operatorname{cosec}^{-1}\left(-\sqrt{2}\right)=-\frac{\pi}{4}}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it combines the reciprocal relationship between cosecant and sine with the principal value concept of an inverse trigonometric function.

The essential conversion is

\[ \operatorname{cosec}y=-\sqrt{2} \]
\[ \frac{1}{\sin y}=-\sqrt{2} \]
\[ \sin y=-\frac{1}{\sqrt{2}} \]

The negative sine value indicates that the principal angle must be negative in the selected principal branch. Therefore, \(-\frac{\pi}{4}\) is obtained.

A very important correction to the given working is that the principal value range of \(\operatorname{cosec}^{-1}x\) is not the entire interval \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The value \(0\) must be excluded because cosecant is undefined at \(0\).

\[ \boxed{ \text{PV range of }\operatorname{cosec}^{-1}x = \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right] } \]
Significance for Competitive Entrance Exams

For competitive entrance examinations, questions involving \(\operatorname{cosec}^{-1}x\) often test whether the student remembers its restricted principal value range. Simply finding an angle with the required sine value is not sufficient; the angle must also belong to the prescribed branch.

In this problem, the standard value

\[ \sin\frac{\pi}{4}=\frac{1}{\sqrt{2}} \]

together with the negative sign gives

\[ \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \]

and the range check confirms that \(-\frac{\pi}{4}\) is the principal value.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. \(\operatorname{cosec}^{-1}x\) gives the principal angle \(y\) satisfying \(\operatorname{cosec}y=x\).

  2. \(\operatorname{cosec}y=\frac{1}{\sin y}\).

  3. The principal value range of \(\operatorname{cosec}^{-1}x\) is \(\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]\).

  4. The value \(0\) is excluded because \(\operatorname{cosec}0\) is undefined.

  5. \(\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\).

  6. \(\sin\left(-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}\).

  7. Therefore, \(\operatorname{cosec}\left(-\frac{\pi}{4}\right)=-\sqrt{2}\).

  8. \(-\frac{\pi}{4}\) lies within the principal value range of \(\operatorname{cosec}^{-1}x\).

  9. Therefore, \(\operatorname{cosec}^{-1}\left(-\sqrt{2}\right)=-\frac{\pi}{4}\).
← Q9
10 / 14  ·  71%
Q11 →
Q11
NUMERIC3 marks
Find the value of the following: \[\tan^{-1}(1)+\cos^{-1}\left(-\frac{1}{2}\right)+\sin^{-1}\left(-\frac{1}{2}\right)\]
📘 Concept & Theory
Concept/Theory
›

When evaluating a sum involving inverse trigonometric functions, each inverse trigonometric term must first be evaluated according to its respective principal value range.

The relevant principal value ranges are:

\[ -\frac{\pi}{2}<\tan^{-1}x<\frac{\pi}{2} \]
\[ 0\leq\cos^{-1}x\leq\pi \]
\[ -\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2} \]

Therefore, we cannot simply choose any angle having the required trigonometric value. The angle must belong to the principal value range of the corresponding inverse function.

The standard values required in this question are:

\[ \tan\frac{\pi}{4}=1 \]
\[ \cos\frac{2\pi}{3}=-\frac{1}{2} \]
\[ \sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2} \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Evaluate \(\tan^{-1}(1)\) using the principal value range of inverse tangent.

  2. Evaluate \(\cos^{-1}\left(-\frac{1}{2}\right)\) using the principal value range of inverse cosine.

  3. Evaluate \(\sin^{-1}\left(-\frac{1}{2}\right)\) using the principal value range of inverse sine.

  4. Substitute all three principal values into the given expression.

  5. Take the LCM of the denominators and simplify the resulting expression.

  6. State the final value.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  25 steps
  1. Step 1: Evaluate \(\tan^{-1}(1)\)
  2. Let
    \[x=\tan^{-1}(1)\]
  3. By the definition of the inverse tangent function,
    \[\tan x=1\]
  4. We know that
    \[\tan\frac{\pi}{4}=1\]
  5. Therefore,
    \[x=\frac{\pi}{4}\]
  6. Since
    \[-\frac{\pi}{2}<\frac{\pi}{4}<\frac{\pi}{2}\]
  7. \(\frac{\pi}{4}\) lies in the principal value range of \(\tan^{-1}x\).
  8. Hence,
    \[\boxed{\tan^{-1}(1)=\frac{\pi}{4}}\]
  9. Step 2: Evaluate \(\cos^{-1}\left(-\frac{1}{2}\right)\)
  10. Let
    \[y=\cos^{-1}\left(-\frac{1}{2}\right)\]
  11. By the definition of the inverse cosine function,
    \[\cos y=-\frac{1}{2}\]
  12. We know that
    \[\cos\frac{\pi}{3}=\frac{1}{2}\]
  13. Since cosine is negative in the second quadrant, the required principal angle is
    \[y=\pi-\frac{\pi}{3}\]
  14. Therefore,
    \[\begin{aligned} y&=\frac{3\pi-\pi}{3}\\&=\frac{2\pi}{3}\end{aligned}\]
  15. Step 3: Evaluate \(\sin^{-1}\left(-\frac{1}{2}\right)\)
  16. Let
    \[z=\sin^{-1}\left(-\frac{1}{2}\right)\]
  17. By the definition of the inverse sine function,
    \[\sin z=-\frac{1}{2}\]
  18. We know that
    \[\sin\frac{\pi}{6}=\frac{1}{2}\]
  19. Since sine is an odd function,
    \[\sin(-\theta)=-\sin\theta\]
  20. Therefore, \sin\left(-\frac{\pi}{6}\right)&=-\sin\frac{\pi}{6}\\&=-\frac{1}{2}\]
  21. Hence,
    \[z=-\frac{\pi}{6}\]
  22. Since
    \[-\frac{\pi}{2} < -\frac{\pi}{6} < \frac{\pi}{2} \]
  23. the value \(-\frac{\pi}{6}\) lies in the principal value range of \(\sin^{-1}x\). Thus,
    \[\boxed{\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}}\]
  24. Step 4: Substitute the Principal Values
  25. The given expression is
    \[\tan^{-1}(1)+\cos^{-1}\left(-\frac{1}{2}\right)+\sin^{-1}\left(-\frac{1}{2}\right)\]
  26. Substituting the values obtained above,
    \[=\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}\]
  27. Step 5: Simplify
  28. The LCM of \(4\), \(3\), and \(6\) is \(12\). Therefore,
    \[\frac{\pi}{4}=\frac{3\pi}{12}\]
    \[\frac{2\pi}{3}=\frac{8\pi}{12}\]
    \[\frac{\pi}{6}=\frac{2\pi}{12}\]
  29. Hence,
    \[\require{cancel}\begin{aligned}\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}&=\frac{3\pi}{12}+\frac{8\pi}{12}-\frac{2\pi}{12}\\ &=\frac{3\pi+8\pi-2\pi}{12}\\&=\frac{\cancelto{3}{9}\pi}{\cancelto{4}{12}}\\&=\frac{3\pi}{4}\end{aligned}\]
  30. Therefore, the required value is
    \[\boxed{\frac{3\pi}{4}}\]
🎯 Exam Significance
Exam Significance
›

This is an important composite principal-value question because it requires students to apply three different inverse trigonometric functions in the same expression.

The most important examination skill is to remember that each inverse function has its own principal value range. In particular:

\[ \tan^{-1}x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]
\[ \cos^{-1}x\in[0,\pi] \]
\[ \sin^{-1}x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \]

The second term is especially important. Since \(\cos^{-1}x\) has its principal values in \([0,\pi]\), the principal value of \(\cos^{-1}\left(-\frac{1}{2}\right)\) is \(\frac{2\pi}{3}\), not \(-\frac{2\pi}{3}\) or another equivalent angle.

Significance for Competitive Entrance Exams

This type of question is useful for competitive entrance examinations because it tests rapid recognition of standard values, correct principal-value selection and algebraic simplification.

A particularly useful strategy is to evaluate each inverse function independently before combining the results. This avoids the common mistake of treating inverse trigonometric functions as ordinary algebraic reciprocals or selecting angles outside their principal ranges.

The three values can be remembered as

\[ \tan^{-1}(1)=\frac{\pi}{4} \]
\[ \cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3} \]
\[ \sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6} \]

After these values are correctly identified, the remaining calculation is straightforward.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. Always evaluate each inverse trigonometric function using its own principal value range.

  2. \(\tan^{-1}(1)=\frac{\pi}{4}\).

  3. \(\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}\).

  4. \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\).

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

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

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

  8. After evaluating the individual inverse functions, substitute their principal values into the original expression.

  9. The final simplification gives \(\frac{3\pi}{4}\).
← Q10
11 / 14  ·  79%
Q12 →
Q12
NUMERIC3 marks
Find the value of the following: \[\cos^{-1}\left(\frac{1}{2}\right)+2\sin^{-1}\left(\frac{1}{2}\right)\]
📘 Concept & Theory
Concept/Theory
›

To evaluate an expression involving inverse trigonometric functions, first determine the principal value of each inverse function separately. The relevant principal value ranges are

\[ \cos^{-1}x\in[0,\pi] \]
\[ \sin^{-1}x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \]

We use the standard trigonometric values

\[ \cos\frac{\pi}{3}=\frac{1}{2} \]
\[ \sin\frac{\pi}{6}=\frac{1}{2} \]

Therefore,

\[ \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \]
\[ \sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \]

Both values satisfy their respective principal value restrictions.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Evaluate \(\cos^{-1}\left(\frac{1}{2}\right)\) using the standard cosine value.

  2. Evaluate \(\sin^{-1}\left(\frac{1}{2}\right)\) using the standard sine value.

  3. Check that both angles lie in their respective principal value ranges.

  4. Substitute the values into the given expression.

  5. Simplify the resulting expression carefully.

  6. State the final answer.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  21 steps
  1. Step 1: Evaluate \(\cos^{-1}\left(\frac{1}{2}\right)\)
  2. Let
    \[x=\cos^{-1}\left(\frac{1}{2}\right)\]
  3. By the definition of the inverse cosine function,
    \[\cos x=\frac{1}{2}\]
  4. We know that
    \[\cos\frac{\pi}{3}=\frac{1}{2}\]
  5. Hence,
    \[x=\frac{\pi}{3}\]
  6. The principal value range of \(\cos^{-1}x\) is
    \[0\leq x\leq\pi\]
  7. Since \(0<\frac{\pi}{3}<\pi\) \(\frac{\pi}{3}\) is an admissible principal value.
  8. Therefore,
    \[\boxed{\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}}\]
  9. Step 2: Evaluate \(\sin^{-1}\left(\frac{1}{2}\right)\)
  10. Let
    \[y=\sin^{-1}\left(\frac{1}{2}\right)\]
  11. By the definition of the inverse sine function,
    \[\sin y=\frac{1}{2}\]
  12. We know that
    \[\sin\frac{\pi}{6}=\frac{1}{2}\]
  13. Hence,
    \[y=\frac{\pi}{6}\]
  14. The principal value range of \(\sin^{-1}x\) is
    \[-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\]
  15. Since
    \[-\frac{\pi}{2}<\frac{\pi}{6}<\frac{\pi}{2}\]
  16. \(\frac{\pi}{6}\) is an admissible principal value. Therefore,
    \[\boxed{\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}}\]
  17. Step 3: Substitute the Values
  18. The given expression is
    \[\cos^{-1}\left(\frac{1}{2}\right)+2\sin^{-1}\left(\frac{1}{2}\right)\]
  19. Substituting the principal values,
    \[=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)\]
  20. Multiplying,
    \[=\frac{\pi}{3}+\frac{2\pi}{6}\]
  21. Since
    \[\frac{2\pi}{6}=\frac{\pi}{3}\]
  22. we get
    \[ =\frac{\pi}{3}+\frac{\pi}{3}\]
  23. Therefore,
    \[\frac{2\pi}{3}\]
  24. Hence, the required value is
    \[ \boxed{\frac{2\pi}{3}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the direct evaluation of two different inverse trigonometric functions in the same expression. The key point is that each inverse function has its own principal value range.

For this problem,

\[ \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \]

whereas

\[ \sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \]

Students should not assume that the same angle represents both inverse functions merely because the argument is the same. The corresponding trigonometric ratios determine different standard angles.

This type of direct-value question is useful for building accuracy before attempting more complicated inverse trigonometric identities and equations.

Significance for Competitive Entrance Exams

For competitive entrance examinations, quick recognition of standard inverse trigonometric values is essential. This question can be solved rapidly once the following two standard values are recalled:

\[ \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \]
\[ \sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \]

The coefficient \(2\) must then be applied to the complete value of \(\sin^{-1}\left(\frac{1}{2}\right)\):

\[ 2\sin^{-1}\left(\frac{1}{2}\right) = 2\left(\frac{\pi}{6}\right) = \frac{\pi}{3} \]

Therefore, the final result follows directly:

\[ \frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3} \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\).

  2. \(\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\).

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

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

  5. Always evaluate each inverse trigonometric function according to its own principal value range.

  6. The coefficient \(2\) multiplies the complete value of \(\sin^{-1}\left(\frac{1}{2}\right)\).

  7. \(2\left(\frac{\pi}{6}\right)=\frac{\pi}{3}\).

  8. Therefore, the required value is \(\frac{2\pi}{3}\).
← Q11
12 / 14  ·  86%
Q13 →
Q13
NUMERIC3 marks
If \(\sin^{-1}x=y,\) then which of the following is correct?
  1. \(0\leq y\leq\pi\)
  2. \(-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\)
  3. \(0
  4. \(-\frac{\pi}{2}
📘 Concept & Theory
Concept/Theory
›

The inverse sine function, \(\sin^{-1}x\), is defined by restricting the domain of the sine function so that it becomes one-one and hence has an inverse.

The principal value range of the inverse sine function is

\[ -\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2} \]

Therefore, if

\[ \sin^{-1}x=y, \]

then necessarily

\[-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\]

The endpoints are included because

\[ \sin\left(-\frac{\pi}{2}\right)=-1 \]

and

\[ \sin\left(\frac{\pi}{2}\right)=1. \]

Thus, the correct interval is a closed interval, not an open interval.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Start with the given relation \(\sin^{-1}x=y\).

  2. Recall the principal value range of the inverse sine function.

  3. Apply that range directly to \(y\).

  4. Compare the resulting interval with the given options.

  5. Select the correct option.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  6 steps
  1. Given — \(\sin^{-1}x=y\)
  2. By the definition of the inverse sine function, \(\sin^{-1}x\) always gives its principal value.
  3. The principal value range of \(\sin^{-1}x\) is
    \[-\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2}\]
  4. Since
    \[y=\sin^{-1}x\]
  5. we can replace \(\sin^{-1}x\) by \(y\):
    \[-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\]
  6. Therefore, the required condition is
    \[\boxed{-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}}\]
  7. Comparing this result with the given options, we find that it corresponds to Option (B).
🎯 Exam Significance
Exam Significance
›

This question directly tests the principal value range of the inverse sine function. Memorising the principal ranges of the six inverse trigonometric functions is essential for Class 12 board examinations.

The most important distinction here is between an open interval and a closed interval:

\[ \boxed{ -\frac{\pi}{2}\leq y\leq\frac{\pi}{2} } \]

The equality signs are important because both \(1\) and \(-1\) belong to the domain of \(\sin^{-1}x\).

Significance for Competitive Entrance Exams

Principal value ranges are frequently used in JEE and other competitive entrance examinations, particularly in questions involving inverse trigonometric identities, equations and composite functions.

A frequent trap is to select \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) instead of \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Checking the endpoint values immediately eliminates this error:

\[ \sin\left(-\frac{\pi}{2}\right)=-1 \]
\[ \sin\left(\frac{\pi}{2}\right)=1 \]

Therefore, both endpoints are valid principal values.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. If \(\sin^{-1}x=y\), then \(y\) is the principal value of the inverse sine function.

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

  3. The interval is closed because both \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) are valid principal values.

  4. \(\sin^{-1}(-1)=-\frac{\pi}{2}\).

  5. \(\sin^{-1}(1)=\frac{\pi}{2}\).

  6. Option (D) is incorrect because it excludes both endpoints.

  7. Option (A) represents the principal value range of \(\cos^{-1}x\).

  8. Therefore, the correct answer is \(\boxed{\text{Option (B)}}\).
← Q12
13 / 14  ·  93%
Q14 →
Q14
NUMERIC3 marks
Find the value of the following: \(\tan^{-1}\sqrt{3}-\sec^{-1}(-2)\)
📘 Concept & Theory
Concept/Theory
›

The principal value range of the inverse tangent function is

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

Therefore, since

\[ \tan\frac{\pi}{3}=\sqrt{3}, \]

we have

\[ \tan^{-1}\sqrt{3}=\frac{\pi}{3}. \]

For the inverse secant function, the principal value range is

\[ [0,\pi]\setminus\left\{\frac{\pi}{2}\right\}. \]

Since

\[ \sec y=-2, \]

we can write

\[ \cos y=-\frac{1}{2}. \]

In the principal range of \(\sec^{-1}x\), the angle having cosine \(-\frac{1}{2}\) is

\[ y=\frac{2\pi}{3}. \]

Hence,

\[ \sec^{-1}(-2)=\frac{2\pi}{3}. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Evaluate \(\tan^{-1}\sqrt{3}\) using the standard tangent value.

  2. Evaluate \(\sec^{-1}(-2)\) by converting secant into cosine.

  3. Apply the principal value range of \(\sec^{-1}x\) to select the correct angle.

  4. Substitute both principal values into the original expression.

  5. Perform the subtraction carefully.

  6. State the final answer and corresponding option.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Step 1: Evaluate \(\tan^{-1}\sqrt{3}\)
  2. Let
    \[x=\tan^{-1}\sqrt{3}\]
  3. By the definition of the inverse tangent function,
    \[\tan x=\sqrt{3}\]
  4. We know the standard trigonometric value
    \[\tan\frac{\pi}{3}=\sqrt{3}\]
  5. Therefore,
    \[x=\frac{\pi}{3}\]
  6. We verify that this angle belongs to the principal value range:
    \[-\frac{\pi}{2}<\frac{\pi}{3}<\frac{\pi}{2}\]
  7. Hence,
    \[\boxed{\tan^{-1}\sqrt{3}=\frac{\pi}{3}}\]
  8. Step 2: Evaluate \(\sec^{-1}(-2)\)
  9. Let
    \[y=\sec^{-1}(-2)\]
  10. By the definition of the inverse secant function,
    \[\sec y=-2\]
  11. Since
    \[\sec y=\frac{1}{\cos y}\]
  12. we have
    \[\frac{1}{\cos y}=-2\]
  13. Taking the reciprocal of both sides
    \[\cos y=-\frac{1}{2}\]
  14. We know that
    \[cos\frac{\pi}{3}=\frac{1}{2}\]
  15. Since the cosine value is negative, the required angle lies in the second quadrant within the principal range of inverse secant.
  16. Therefore,
    \[y=\pi-\frac{\pi}{3}\]
  17. Simplifying,
    \[\begin{aligned}y&=\frac{3\pi-\pi}{3}\\&=\frac{2\pi}{3}\end{aligned}\]
  18. Step 3: Substitute the Principal Values
  19. The original expression is
    \[\tan^{-1}\sqrt{3}-\sec^{-1}(-2)\]
  20. Substituting the values obtained above,
    \[=\frac{\pi}{3}-\frac{2\pi}{3}\]
  21. Step 4: Simplify
  22. Taking the common denominator \(3\),
    \[\begin{aligned}\frac{\pi}{3}-\frac{2\pi}{3}&=\frac{\pi-2\pi}{3}\\&=-\frac{\pi}{3}\end{aligned}\]
  23. Therefore, the required value is
    \[\boxed{-\frac{\pi}{3}}\]
🎯 Exam Significance
Exam Significance
›

This question tests two important skills: evaluating standard inverse trigonometric values and correctly handling the principal value range of the inverse secant function.

The critical step is

\[ \sec^{-1}(-2)=\frac{2\pi}{3}, \]

because the principal value of inverse secant is selected from

\[ [0,\pi]\setminus\left\{\frac{\pi}{2}\right\}. \]

A common mistake is to choose \(-\frac{2\pi}{3}\) merely because its secant is also \(-2\). However, \(-\frac{2\pi}{3}\) does not belong to the prescribed principal value range.

Significance for Competitive Entrance Exams

This is a useful competitive-exam question because it tests whether the student can distinguish between an inverse trigonometric function and the complete set of angles satisfying the corresponding trigonometric equation.

A fast method is:

\[ \tan^{-1}\sqrt{3}=\frac{\pi}{3} \]
\[ \sec^{-1}(-2) = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \]

Therefore,

\[ \frac{\pi}{3}-\frac{2\pi}{3} = -\frac{\pi}{3}. \]

This approach is particularly efficient in multiple-choice questions.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. \(\tan^{-1}\sqrt{3}=\frac{\pi}{3}\).

  2. \(\sec^{-1}(-2)=\frac{2\pi}{3}\).

  3. \(\sec^{-1}x\) has principal value range \([0,\pi]\setminus\left\{\frac{\pi}{2}\right\}\).

  4. \(\sec^{-1}(-2)\) can be evaluated by converting \(\sec y=-2\) into \(\cos y=-\frac{1}{2}\).

  5. The original textbook expression contains a minus sign between the two inverse functions.

  6. Therefore, \(\frac{\pi}{3}-\frac{2\pi}{3}=-\frac{\pi}{3}\).

  7. The correct multiple-choice option is (B).
← Q13
14 / 14  ·  100%
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NCERT Class 12 Maths Ex 2.1 Solutions | Inverse Trig
NCERT Class 12 Maths Ex 2.1 Solutions | Inverse Trig — Complete Notes & Solutions · academia-aeternum.com
Special attention is given to the principal value ranges of inverse trigonometric functions, standard trigonometric values, quadrant selection and common examination mistakes. These NCERT solutions are useful for CBSE Class 12 Board Exams, JEE and other competitive entrance examinations. The detailed approach also builds a strong foundation for solving inverse trigonometric identities, equations and more advanced problems. Use these Exercise 2.1 solutions for systematic revision, concept…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    ✔️ True / False

    Frequently Asked Questions

    Exercise 2.1 of NCERT Class 12 Mathematics Chapter 2, Inverse Trigonometric Functions, focuses on finding principal values of inverse trigonometric functions and evaluating expressions involving them.

    The principal value range of sin?¹x is -p/2 = sin?¹x = p/2.

    The principal value range of cos?¹x is 0 = cos?¹x = p.

    The principal value range of tan?¹x is -p/2 < tan?¹x < p/2.

    The principal value range of sec?¹x is [0, p] excluding p/2.

    The principal value range of cosec?¹x is [-p/2, 0) ? (0, p/2].

    First identify the corresponding trigonometric ratio and standard angle, then select the angle that lies within the principal value range of the relevant inverse trigonometric function.

    The value is 3p/4 because tan?¹(1) = p/4, cos?¹(-1/2) = 2p/3, and sin?¹(-1/2) = -p/6.

    The value is 2p/3 because cos?¹(1/2) = p/3 and sin?¹(1/2) = p/6.

    The value is -p/3 because tan?¹v3 = p/3 and sec?¹(-2) = 2p/3.

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