NCERT Class 12 Maths Ex 2.2 Solutions

📘 Concept & Theory

Concept/Theory

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We use the standard trigonometric identity

\[ \sin 3\theta=3\sin\theta-4\sin^3\theta \]

The principal value of $\sin^{-1}x$ always lies in the interval

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

However, while simplifying an expression of the form $\sin^{-1}(\sin 3\theta)$, we cannot directly write $\sin^{-1}(\sin 3\theta)=3\theta$ for every value of $\theta$. This is true only when $3\theta$ lies within the principal range of $\sin^{-1}$, namely

\[ -\frac{\pi}{2}\leq3\theta\leq\frac{\pi}{2}. \]

Therefore,

\[ -\frac{\pi}{6}\leq\theta\leq\frac{\pi}{6}. \]

Since $x=\sin\theta$, this corresponds to

\[ -\frac{1}{2}\leq x\leq\frac{1}{2}. \]

Thus, the stated identity is valid on the principal-value interval

\[ \boxed{-\frac12\leq x\leq\frac12}. \]

🗺️ Solution Roadmap

Step-by-step Plan

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Put $x=\sin\theta$, so that $\theta=\sin^{-1}x$.
Use the identity $\sin3\theta=3\sin\theta-4\sin^3\theta$.
Express $3x-4x^3$ in terms of $\sin3\theta$.
Apply $\sin^{-1}$ to both sides.
Check the principal-value condition for $\sin^{-1}(\sin3\theta)$.
Substitute $\theta=\sin^{-1}x$ and obtain the required result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Let $x=\sin\theta$
Then, by taking the inverse sine of both sides,
\[\theta=\sin^{-1}x\]
Since $\sin^{-1}x$ has its principal value in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, we take
\[-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\]
To simplify $\sin^{-1}(\sin3\theta)$ directly as $3\theta$, we additionally require
\[-\frac{\pi}{2}\leq3\theta\leq\frac{\pi}{2}\]
Dividing throughout by $3$,
\[-\frac{\pi}{6}\leq\theta\leq\frac{\pi}{6}\]
Therefore,
\[-\frac12\leq\sin\theta\leq\frac12\]
Since $x=\sin\theta$, we have
\[-\frac12\leq x\leq\frac12\]
Now use the triple-angle identity
\[\sin3\theta=3\sin\theta-4\sin^3\theta\]
Substituting $x=\sin\theta$, we get
\[\sin3\theta=3x-4x^3\]
Hence,
\[3x-4x^3=\sin3\theta\]
Consider the right-hand side of the required identity:
\[\begin{aligned}\sin^{-1}\left(3x-4x^3\right)&=\sin^{-1}\left(\sin3\theta\right)\\&=3\theta,\end{aligned}\]
where the last step is valid because
\[-\frac{\pi}{2}\leq3\theta\leq\frac{\pi}{2}\]
But
\[\theta=\sin^{-1}x\]
Therefore,
\[3\theta=3\sin^{-1}x\]
Consequently,
\[\begin{aligned}\sin^{-1}\left(3x-4x^3\right)&=3\theta\\&=3\sin^{-1}x\end{aligned}\]
Hence,
\[\boxed{3\sin^{-1}x=\sin^{-1}\left(3x-4x^3\right)}\]
for
\[\boxed{-\frac12\leq x\leq\frac12}\]
Important Principal-Value Note
The step
\[\sin^{-1}(\sin3\theta)=3\theta\]
must not be used without checking the principal range. In general, $\sin^{-1}(\sin y)=y$ only when
\[ -\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\]
This restriction is essential in inverse-trigonometric problems and is a common source of errors in board examinations and competitive entrance tests.

🎯 Exam Significance

Exam Significance

›

This problem tests the use of the triple-angle identity for sine.
It tests the relationship between a trigonometric function and its inverse function.
The principal-value restriction of $\sin^{-1}x$ is an important scoring point.
Writing the range condition explicitly prevents an otherwise incomplete proof.
The substitution $x=\sin\theta$ is a standard technique for simplifying inverse-trigonometric expressions.

Significance for Competitive Entrance Exams

Questions involving $\sin^{-1}(\sin\theta)$ frequently test principal-value concepts.
The identity $ \sin3\theta=3\sin\theta-4\sin^3\theta $ is useful for simplifying higher-degree polynomial expressions in $x$.
Domain and range restrictions can determine whether an apparently familiar identity is actually valid.
Careful range analysis is particularly important in multiple-choice questions, where an omitted restriction can lead to an incorrect option.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

Use the substitution $x=\sin\theta$ when an expression contains $\sin^{-1}x$.
Remember the identity
\[ \sin3\theta=3\sin\theta-4\sin^3\theta. \]
The principal range of $\sin^{-1}x$ is
\[ -\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2}. \]
For $\sin^{-1}(\sin3\theta)=3\theta$, we must have
\[ -\frac{\pi}{2}\leq3\theta\leq\frac{\pi}{2}. \]
This gives
\[ -\frac{\pi}{6}\leq\theta\leq\frac{\pi}{6}. \]
Consequently, the identity holds for
\[ \boxed{-\frac12\leq x\leq\frac12}. \]
Never simplify $\sin^{-1}(\sin y)$ to $y$ without checking whether $y$ belongs to the principal range of $\sin^{-1}$.

📘 Concept & Theory

Concept/Theory

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This problem is based on the triple-angle identity for cosine:

\[ \cos3\theta=4\cos^3\theta-3\cos\theta \]

When an expression contains $\cos^{-1}x$, a useful substitution is

\[ x=\cos\theta \]

which gives

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

The principal value of $\cos^{-1}x$ lies in the interval

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

However, while simplifying $\cos^{-1}(\cos3\theta)$, we cannot always write $\cos^{-1}(\cos3\theta)=3\theta$. This direct simplification is valid only when $3\theta$ lies within the principal range of $\cos^{-1}$, namely

\[ 0\leq3\theta\leq\pi. \]

Therefore,

\[ 0\leq\theta\leq\frac{\pi}{3}. \]

Since $x=\cos\theta$ and cosine is decreasing on $[0,\pi]$, we obtain

\[ \frac12\leq x\leq1. \]

Hence, the identity is valid in the principal-value sense for

\[ \boxed{\frac12\leq x\leq1}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Put $x=\cos\theta$.
Therefore, $\theta=\cos^{-1}x$.
Rewrite $4x^3-3x$ in terms of $\cos\theta$.
Apply the triple-angle identity
\[ \cos3\theta=4\cos^3\theta-3\cos\theta. \]
Convert the resulting expression into $\cos^{-1}(\cos3\theta)$.
Check the principal-value condition for $\cos^{-1}$.
Substitute $\theta=\cos^{-1}x$ to obtain the required result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Let $x=\cos\theta$
Taking inverse cosine on both sides,
\[\theta=\cos^{-1}x\]
Since the principal range of $\cos^{-1}x$ is
\[0\leq\cos^{-1}x\leq\pi\]
we have
\[0\leq\theta\leq\pi\]
Now consider the right-hand side:
\[\cos^{-1}\left(4x^3-3x\right)\]
Substituting $x=\cos\theta$, we get
\[\cos^{-1}\left(4\cos^3\theta-3\cos\theta\right)\]
Using the triple-angle identity
\[\cos3\theta=4\cos^3\theta-3\cos\theta,\]
we obtain
\[4\cos^3\theta-3\cos\theta=\cos3\theta\]
Therefore,
\[\begin{aligned}\cos^{-1}\left(4\cos^3\theta-3\cos\theta\right)&=\cos^{-1}(\cos3\theta)\end{aligned}\]
Now we must consider the principal-value condition carefully.
The principal range of $\cos^{-1}y$ is
\[0\leq\cos^{-1}y\leq\pi\]
Hence,
\[\cos^{-1}(\cos3\theta)=3\theta\]
only when
\[ 0\leq3\theta\leq\pi\]
Dividing throughout by $3$,
\[0\leq\theta\leq\frac{\pi}{3}\]
Since $x=\cos\theta$, and cosine is decreasing in $[0,\pi]$, we get
\[\cos\frac{\pi}{3}\leq\cos\theta\leq\cos0\]
Therefore,
\[\frac12\leq x\leq1\]
Under this condition,
\[\begin{aligned}\cos^{-1}\left(4x^3-3x\right)&=\cos^{-1}(\cos3\theta)\\&=3\theta\end{aligned}\]
But
\[\theta=\cos^{-1}x\]
Therefore,
\[3\theta=3\cos^{-1}x\]
Hence,
\[\begin{aligned}\cos^{-1}\left(4x^3-3x\right)&=3\theta\\&=3\cos^{-1}x\end{aligned}\]
Thus,
\[\boxed{3\cos^{-1}x=\cos^{-1}\left(4x^3-3x\right)}\]
for
\[\boxed{\frac12\leq x\leq1}\]
Why the Range Condition Matters
A common mistake is to write
\[\cos^{-1}(\cos3\theta)=3\theta\]
without checking the principal range. This is not universally true. The inverse cosine function returns values only in $[0,\pi]$
Therefore, $3\theta$ must belong to this interval for the direct simplification to be valid:
\[0\leq3\theta\leq\pi\]
This gives
\[ 0\leq\theta\leq\frac{\pi}{3}\]
and consequently
\[ \boxed{\frac12\leq x\leq1}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the application of the cosine triple-angle identity.
It tests the correct use of substitution $x=\cos\theta$.
It checks understanding of the principal range of $\cos^{-1}x$.
Writing the range restriction makes the proof mathematically complete.
The question is useful for testing whether students can distinguish a trigonometric identity from an inverse-trigonometric identity.

Significance for Competitive Entrance Exams

Principal-value questions involving $\cos^{-1}(\cos\theta)$ are frequently used to test conceptual understanding.
The polynomial $4x^3-3x$ should immediately suggest the identity $\cos3\theta=4\cos^3\theta-3\cos\theta$.
Domain and range restrictions can be decisive in multiple-choice questions.
Recognising standard multiple-angle polynomials can significantly reduce calculation time.
Careful handling of inverse functions prevents the common error of treating them as ordinary reciprocal functions.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 8 points

For expressions containing $\cos^{-1}x$, the substitution $x=\cos\theta$ is often effective.
The important triple-angle identity is
\[ \cos3\theta=4\cos^3\theta-3\cos\theta. \]
The principal range of $\cos^{-1}x$ is
\[ 0\leq\cos^{-1}x\leq\pi. \]
The relation
\[ \cos^{-1}(\cos y)=y \]
is directly valid only when
\[ 0\leq y\leq\pi. \]
For $y=3\theta$, we require
\[ 0\leq3\theta\leq\pi. \]
Hence,
\[ 0\leq\theta\leq\frac{\pi}{3}. \]
Since $x=\cos\theta$,
\[ \frac12\leq x\leq1. \]
The final identity is
\[ \boxed{3\cos^{-1}x=\cos^{-1}\left(4x^3-3x\right)} \]
for
\[ \boxed{\frac12\leq x\leq1}. \]

📘 Concept & Theory

Concept/Theory

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The expression contains both $\sqrt{1+x^2}$ and $x$, which suggests the substitution

\[x=\tan\theta\]

This substitution is useful because

\[1+\tan^2\theta=\sec^2\theta\]

The resulting expression can then be transformed using the half-angle identity

\[\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}\]

Another useful equivalent form is

\[\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}\]

Since the original expression contains division by $x$, we require

\[\boxed{x\neq0}\]

🗺️ Solution Roadmap

Step-by-step Plan

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Put $x=\tan\theta$.
Therefore, $\theta=\tan^{-1}x$.
Use $1+\tan^2\theta=\sec^2\theta$ to simplify the square root.
Rewrite the resulting expression in terms of $\sin\theta$ and $\cos\theta$.
Apply the half-angle identity to show that the expression equals $\tan\frac{\theta}{2}$.
Apply $\tan^{-1}$ to both sides.
Substitute $\theta=\tan^{-1}x$ to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 20 steps

Let
\[x=\tan\theta\]
Taking inverse tangent on both sides,
\[\theta=\tan^{-1}x\]
Since the principal range of $\tan^{-1}x$ is
\[-\frac{\pi}{2}<\theta<\frac{\pi}{2}\]
we have
\[\cos\theta>0\]
Now consider the expression inside the inverse tangent:
\[\frac{\sqrt{1+x^2}-1}{x}\]
Substituting $x=\tan\theta$, we get
\[\begin{aligned}\frac{\sqrt{1+x^2}-1}{x}&=\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\end{aligned}\]
Using
\[1+\tan^2\theta=\sec^2\theta\]
we obtain
\[\begin{aligned}\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}&=\frac{\sqrt{\sec^2\theta}-1}{\tan\theta}\end{aligned}\]
Since $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$, we have $\cos\theta>0$, and hence $\sec\theta>0$. Therefore,
\[\sqrt{\sec^2\theta}=\sec\theta\]
Thus,
\[\begin{aligned}\frac{\sqrt{1+x^2}-1}{x}&=\frac{\sec\theta-1}{\tan\theta}\end{aligned}\]
Now express $\sec\theta$ and $\tan\theta$ in terms of sine and cosine:
\[\begin{aligned}\frac{\sec\theta-1}{\tan\theta}&=\frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}\\ &=\frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}\\ &=\frac{1-\cos\theta}{\sin\theta}\end{aligned}\]
Using the half-angle identities
\[1-\cos\theta=2\sin^2\frac{\theta}{2}\quad\text{and}\]

\[\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\]
we get
\[\begin{aligned}\frac{1-\cos\theta}{\sin\theta}&=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\ &=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\\ &=\tan\frac{\theta}{2}\end{aligned}\]
Therefore,
\[\boxed{\frac{\sqrt{1+x^2}-1}{x}=\tan\frac{\theta}{2}}\]
Hence, the original expression becomes
\[\begin{aligned}\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)&=\tan^{-1}\left(\tan\frac{\theta}{2}\right)\end{aligned}\]
Since
\[-\frac{\pi}{2}<\theta<\frac{\pi}{2}\]
we have
\[-\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}\]
Thus, $\frac{\theta}{2}$ lies within the principal range of $\tan^{-1}x$, namely
\[-\frac{\pi}{2} < y<\frac{\pi}{2}\]
Therefore,
\[\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}\]
But
\[\theta=\tan^{-1}x\]
Therefore,
\[\frac{\theta}{2}=\frac12\tan^{-1}x\]
Hence, the simplified result is
\[\boxed{\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac12\tan^{-1}x}\]
where $\boxed{x\neq0}$

🎯 Exam Significance

Exam Significance

›

This problem combines substitution, fundamental trigonometric identities and half-angle identities.
It tests whether students understand the principal range of the inverse tangent function.
The step involving $\sqrt{\sec^2\theta}$ requires attention to the sign of $\sec\theta$.
Showing the intermediate conversion to $\frac{1-\cos\theta}{\sin\theta}$ makes the half-angle application clear and earns method marks.
The rationalisation method provides an efficient alternative that can be useful when a shorter solution is required.

Significance for Competitive Entrance Exams

The substitution $x=\tan\theta$ converts a radical expression into a standard trigonometric form.
Recognising
\[ 1+\tan^2\theta=\sec^2\theta \]
can substantially reduce the calculation.
The expression
\[ \frac{1-\cos\theta}{\sin\theta} \]
should immediately suggest
\[ \tan\frac{\theta}{2}. \]
Rationalisation can provide a fast route to the same half-angle form.
Principal-value restrictions are essential when applying inverse trigonometric functions in objective questions.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

For
\[ \sqrt{1+x^2}, \]
the substitution $x=\tan\theta$ is particularly useful.
Use
\[ 1+\tan^2\theta=\sec^2\theta. \]
Since
\[ -\frac{\pi}{2}<\theta<\frac{\pi}{2}, \]
we have
\[ \sqrt{\sec^2\theta}=\sec\theta. \]
The transformation
\[ \frac{\sec\theta-1}{\tan\theta} = \frac{1-\cos\theta}{\sin\theta} \]
is the key intermediate step.
Using the half-angle identities,
\[ \frac{1-\cos\theta}{\sin\theta} = \tan\frac{\theta}{2}. \]
Therefore,
\[ \boxed{ \tan^{-1}\left( \frac{\sqrt{1+x^2}-1}{x} \right) = \frac12\tan^{-1}x }. \]
The original expression requires
\[ \boxed{x\neq0}. \]

📘 Concept & Theory

Concept/Theory

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This problem is based on the standard half-angle identities:

\[ 1-\cos x=2\sin^2\frac{x}{2} \]

and

\[ 1+\cos x=2\cos^2\frac{x}{2}. \]

Therefore,

\[ \frac{1-\cos x}{1+\cos x} = \tan^2\frac{x}{2}. \]

However, when taking the square root, we must remember that $\sqrt{a^2}=|a|$. Hence,

\[ \sqrt{\tan^2\frac{x}{2}} = \left|\tan\frac{x}{2}\right|. \]

This absolute-value step is essential. The square-root function always gives a non-negative value.

Also, the denominator $1+\cos x$ must not be zero. Therefore,

\[ 1+\cos x\neq0 \]

which gives

\[ x\neq(2n+1)\pi,\qquad n\in\mathbb Z. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Start with the expression inside the square root.
Use the half-angle identities for $1-\cos x$ and $1+\cos x$.
Reduce the fraction to $\tan^2\frac{x}{2}$.
Take the square root carefully using
\[ \sqrt{a^2}=|a|. \]
Apply the inverse tangent function.
For the standard interval $0\leq x<\pi$, remove the absolute value and obtain the simplified answer $\frac{x}{2}$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Consider
\[\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}\]
Using the half-angle identity
\[1-\cos x=2\sin^2\frac{x}{2}\quad\text{and}\]

\[1+\cos x=2\cos^2\frac{x}{2}\]
we obtain
\[\begin{aligned}\frac{1-\cos x}{1+\cos x}&=\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}\\ &=\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}\\ &=\tan^2\frac{x}{2}\end{aligned}\]
Therefore,
\[\sqrt{\frac{1-\cos x}{1+\cos x}}=\sqrt{\tan^2\frac{x}{2}}\]
Using
\[\sqrt{a^2}=|a|\]
we get
\[\boxed{\sqrt{\frac{1-\cos x}{1+\cos x}}=\left|\tan\frac{x}{2}\right|}\]
Hence,
\[\begin{aligned}\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}&=\tan^{-1}\left|\tan\frac{x}{2}\right|\end{aligned}\]
For the Principal Interval $0\leq x<\pi$
If
\[0\leq x<\pi\]
then
\[0\leq\frac{x}{2}<\frac{\pi}{2}\]
In this interval, $\tan\frac{x}{2}\geq0$. Therefore,
\[\left|\tan\frac{x}{2}\right|=\tan\frac{x}{2}\]
Thus,
\[\begin{aligned}\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}&=\tan^{-1}\left(\tan\frac{x}{2}\right)\\ =\frac{x}{2}\end{aligned}\]
Hence, for
\[\boxed{0\leq x<\pi}\]
the simplified result is
\[\boxed{\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}=\frac{x}{2}}\]
General Form
Without imposing the additional restriction $0\leq x<\pi$, the correct simplification is
\[\boxed{\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}=\tan^{-1}\left|\tan\frac{x}{2}\right|}\]
where $x\neq(2n+1)\pi,\qquad n\in\mathbb Z$
Thus, the commonly written answer
\[\frac{x}{2}\]
requires an appropriate restriction on $x$. For the standard principal interval $0\leq x<\pi$, it is directly valid.

🎯 Exam Significance

Exam Significance

›

This problem tests the use of half-angle identities.
It tests the correct treatment of square roots involving squares.
The identity
\[ \sqrt{a^2}=|a| \]
is an important algebraic principle and should not be omitted.
Students should state the relevant interval before replacing
\[ \tan^{-1}\left(\tan\frac{x}{2}\right) \]
by $\frac{x}{2}$.
Writing the intermediate step involving the absolute value makes the solution mathematically rigorous.

Significance for Competitive Entrance Exams

Half-angle identities can convert complicated-looking inverse-trigonometric expressions into simple angular forms.
The expression
\[ \frac{1-\cos x}{1+\cos x} \]
should immediately suggest the identity
\[ \tan^2\frac{x}{2}. \]
The absolute-value issue can be used to create conceptual traps in objective questions.
Principal-range analysis is essential when simplifying inverse-trigonometric functions.
Recognising domain restrictions quickly can eliminate incorrect options without lengthy calculations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use
\[ 1-\cos x=2\sin^2\frac{x}{2} \]
and
\[ 1+\cos x=2\cos^2\frac{x}{2}. \]
Therefore,
\[ \frac{1-\cos x}{1+\cos x} = \tan^2\frac{x}{2}. \]
Always remember
\[ \sqrt{a^2}=|a|, \]
so
\[ \sqrt{\tan^2\frac{x}{2}} = \left|\tan\frac{x}{2}\right|. \]
The completely general simplification is
\[ \boxed{ \tan^{-1}\left|\tan\frac{x}{2}\right| }. \]
For the standard interval
\[ 0\leq x<\pi, \]
we have
\[ \left|\tan\frac{x}{2}\right| = \tan\frac{x}{2}. \]
Hence, for $0\leq x<\pi$,
\[ \boxed{ \tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}} = \frac{x}{2} }. \]
The original expression is undefined when
\[ 1+\cos x=0, \]
i.e.
\[ \boxed{x=(2n+1)\pi,\quad n\in\mathbb Z}. \]

📘 Concept & Theory

Concept/Theory

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The expression can be simplified using the tangent subtraction formula:

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

By dividing the numerator and denominator by $\cos x$, the given expression can be converted into

\[ \frac{1-\tan x}{1+\tan x}. \]

Since

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

we can write

\[ \frac{1-\tan x}{1+\tan x} = \frac{\tan\frac{\pi}{4}-\tan x} {1+\tan\frac{\pi}{4}\tan x}. \]

Therefore,

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

However, an important point must be considered before applying $\tan^{-1}$. The principal range of $\tan^{-1}y$ is

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

Hence,

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

only when

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

For other values of $\theta$, an appropriate multiple of $\pi$ must be added or subtracted.

🗺️ Solution Roadmap

Step-by-step Plan

›

Divide the numerator and denominator by $\cos x$.
Convert the expression into $\frac{1-\tan x}{1+\tan x}$.
Use $\tan\frac{\pi}{4}=1$.
Apply the tangent subtraction formula.
Obtain
\[ \tan\left(\frac{\pi}{4}-x\right). \]
Apply $\tan^{-1}$, taking its principal range into account.
State the simplified result with the appropriate interval for $x$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 24 steps

Consider
\[\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\]
Divide the numerator and denominator inside the fraction by $\cos x$
\[\begin{aligned}\frac{\cos x-\sin x}{\cos x+\sin x}&=\frac{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}\\ &=\frac{1-\tan x}{1+\tan x}\end{aligned}\]
Since
\[ \tan\frac{\pi}{4}=1\]
we can write
\[\begin{aligned}\frac{1-\tan x}{1+\tan x}&=\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\end{aligned}\]
Using
\[\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B},\]
with $A=\frac{\pi}{4},\qquad B=x,$
we obtain
\[ \frac{1-\tan x}{1+\tan x} = \tan\left(\frac{\pi}{4}-x\right). \]
Therefore,
\[\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\tan^{-1}\left[\tan\left(\frac{\pi}{4}-x\right)\right]\]
Principal-Value Analysis
Let
\[\theta=\frac{\pi}{4}-x\]
The principal range of $\tan^{-1}$ is
\[-\frac{\pi}{2}<\tan^{-1}y<\frac{\pi}{2}\]
Therefore, if
\[-\frac{\pi}{2}<\theta<\frac{\pi}{2}\]
then
\[\tan^{-1}(\tan\theta)=\theta\]
Substituting
\[\theta=\frac{\pi}{4}-x\]
we require
\[ -\frac{\pi}{2} < \frac{\pi}{4}-x < \frac{\pi}{2} \]
Solving the left inequality
\[-\frac{\pi}{2} < \frac{\pi}{4}-x\]

\[x < \frac{3\pi}{4}\]
Solving the right inequality
\[\frac{\pi}{4}-x < \frac{\pi}{2}\]

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

\[x >-\frac{\pi}{4}\]
Hence,
\[-\frac{\pi}{4} < x < \frac{3\pi}{4}\]
Within this interval, the direct simplification is valid:
\[\boxed{\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi}{4}-x}\]
General Result
For arbitrary real $x$, the expression is not simply $\frac{\pi}{4}-x$, because $\tan^{-1}$ returns only its principal value.
Since
\[\tan^{-1}(\tan\theta)=\theta-k\pi,\quad k\in\mathbb Z\]
where $k$ is chosen so that the result lies in
\[\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\]
the general answer can be expressed as
\[\boxed{\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi}{4}-x-k\pi}\]
where $k\in\mathbb Z$ is chosen to make the result lie in the principal range of $\tan^{-1}$
The original expression is undefined whenever
\[\cos x+\sin x=0\]
This occurs when
\[\tan x=-1\]
so
\[\boxed{x=-\frac{\pi}{4}+n\pi,\qquad n\in\mathbb Z}\]
Piecewise Form on $0\leq x<\pi$
If the question is considered over the commonly used interval $0\leq x<\pi$, the principal-value answer is
\[\boxed{\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\begin{cases}\frac{\pi}{4}-x,&0\leq x < \frac{3\pi}{4},\\[6pt] \frac{5\pi}{4}-x,&\frac{3\pi}{4} < x < \pi.\end{cases}}\]
The point $x=\frac{3\pi}{4}$ is excluded because
\[\cos\frac{3\pi}{4}+\sin\frac{3\pi}{4}=0\]

🎯 Exam Significance

Exam Significance

›

This problem tests the tangent subtraction formula and its application to inverse trigonometric functions.
It is important to write the correct denominator $1+\tan A\tan B$.
The principal range of $\tan^{-1}$ is an essential part of the solution.
Students should not automatically cancel $\tan^{-1}$ with $\tan$ without checking the range.
Stating the domain restriction makes the solution complete and avoids loss of conceptual marks.

Significance for Competitive Entrance Exams

The form
\[ \frac{1-\tan x}{1+\tan x} \]
should immediately suggest the identity for $\tan(A-B)$.
Recognising $\tan\frac{\pi}{4}=1$ makes the transformation rapid.
Principal-value corrections are frequently tested in multiple-choice and assertion-reasoning questions.
The denominator condition
\[ \cos x+\sin x\neq0 \]
can be used to identify excluded values quickly.
For arbitrary $x$, the result must be adjusted by an integer multiple of $\pi$ to remain within the principal range.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use
\[ \tan(A-B) = \frac{\tan A-\tan B} {1+\tan A\tan B}. \]
Divide numerator and denominator by $\cos x$:
\[ \frac{\cos x-\sin x}{\cos x+\sin x} = \frac{1-\tan x}{1+\tan x}. \]
Since
\[ \tan\frac{\pi}{4}=1, \]
we obtain
\[ \frac{1-\tan x}{1+\tan x} = \tan\left(\frac{\pi}{4}-x\right). \]
Do not automatically use
\[ \tan^{-1}(\tan\theta)=\theta. \]
This requires
\[ -\frac{\pi}{2} < \theta<\frac{\pi}{2}. \]
For
\[ -\frac{\pi}{4} < x < \frac{3\pi}{4}, \]
the simplified result is
\[ \boxed{\frac{\pi}{4}-x} \]
For arbitrary real $x$, the result must be adjusted by an appropriate multiple of $\pi$.
The original expression is undefined when
\[ \boxed{x=-\frac{\pi}{4}+n\pi,\qquad n\in\mathbb Z} \]

📘 Concept & Theory

Concept/Theory

›

The expression contains the radical

\[ \sqrt{a^2-x^2}. \]

This form suggests the trigonometric substitution

\[ x=a\sin\theta. \]

This is useful because

\[ a^2-x^2=a^2-a^2\sin^2\theta=a^2(1-\sin^2\theta)=a^2\cos^2\theta\]

Under the standard assumption $a>0$ and the principal choice

\[ -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}, \]

we have $\cos\theta\geq0$. Therefore,

\[ \sqrt{a^2\cos^2\theta}=a\cos\theta. \]

The expression then reduces to $\tan\theta$, after which the inverse tangent function gives $\theta$.

Domain of the Given Expression

Since the denominator contains $\sqrt{a^2-x^2}$, we require

\[ a^2-x^2>0 \]

Thus, assuming $a>0$,

\[ -a < x < a \]

Therefore,

\[ \boxed{-a < x < a} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Use the substitution $x=a\sin\theta$.
Determine $\theta$ in terms of $x$:
\[ \theta=\sin^{-1}\left(\frac{x}{a}\right). \]
Substitute $x=a\sin\theta$ into the radical.
Use $1-\sin^2\theta=\cos^2\theta$.
Simplify the resulting fraction to $\tan\theta$.
Apply $\tan^{-1}$ and use the principal range of $\theta$.
Substitute the value of $\theta$ to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 17 steps

Assume $a > 0$
Let
\[x=a\sin\theta\]
Dividing by $a$, we get
\[\frac{x}{a}=\sin\theta\]
Taking inverse sine on both sides,
\[\theta=\sin^{-1}\left(\frac{x}{a}\right)\]
Since the principal range of $\sin^{-1}$ is
\[-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\]
we have
\[\cos\theta\geq0\]
Now consider the expression inside the inverse tangent:
\[\frac{x}{\sqrt{a^2-x^2}}\]
Substituting $x=a\sin\theta$,
\[\begin{aligned}\frac{x}{\sqrt{a^2-x^2}}&=\frac{a\sin\theta}{\sqrt{a^2-a^2\sin^2\theta}}\end{aligned}\]
Taking $a^2$ common inside the square root,
\[\begin{aligned}\frac{x}{\sqrt{a^2-x^2}}&=\frac{a\sin\theta}{\sqrt{a^2(1-\sin^2\theta)}}\end{aligned}\]
Using
\[1-\sin^2\theta=\cos^2\theta\]
we obtain
\[\begin{aligned}\frac{x}{\sqrt{a^2-x^2}}&=\frac{a\sin\theta}{\sqrt{a^2\cos^2\theta}}\end{aligned}\]
Since $a>0$ and $\cos\theta\geq0$
\[\sqrt{a^2\cos^2\theta}=a\cos\theta\]
Therefore,
\[\begin{aligned}\frac{x}{\sqrt{a^2-x^2}}&=\frac{a\sin\theta}{a\cos\theta}\\ &=\frac{\sin\theta}{\cos\theta}\\ &=\tan\theta\end{aligned}\]
Hence, the original expression becomes
\[\begin{aligned}\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)&=\tan^{-1}(\tan\theta)\end{aligned}\]
Since
\[-\frac{\pi}{2}<\theta<\frac{\pi}{2},\]
for the given domain $-a < x < a$, $\theta$ lies in the principal range of $\tan^{-1}$. Therefore,
\[\tan^{-1}(\tan\theta)=\theta\]
Thus,
\[\begin{aligned}\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)&=\theta\\ &=\sin^{-1}\left(\frac{x}{a}\right)\end{aligned}\]
Hence, the required result is
\[\boxed{\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin^{-1}\left(\frac{x}{a}\right)}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the use of a suitable trigonometric substitution for a radical expression.
The substitution $x=a\sin\theta$ is a standard technique for expressions involving $a^2-x^2$.
It tests the identities
\[ 1-\sin^2\theta=\cos^2\theta \]
and
\[ \tan\theta=\frac{\sin\theta}{\cos\theta}. \]
Correctly handling the square root and the principal range demonstrates conceptual understanding.
The final answer connects inverse tangent and inverse sine functions, making the problem particularly relevant to the inverse-trigonometric-functions chapter.

Significance for Competitive Entrance Exams

The substitution $x=a\sin\theta$ should be recognised immediately when $a^2-x^2$ occurs under a square root.
The expression has a direct right-triangle interpretation:
\[ \sin\theta=\frac{x}{a}, \quad \tan\theta=\frac{x}{\sqrt{a^2-x^2}}. \]
This type of identity is useful for quickly simplifying inverse-trigonometric expressions in objective questions.
Domain restrictions can be used to verify whether a proposed answer is valid.
The result is an important conversion identity between $\tan^{-1}$ and $\sin^{-1}$.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

For a radical of the form
\[ \sqrt{a^2-x^2}, \]
use
\[ x=a\sin\theta. \]
The substitution gives
\[ \theta=\sin^{-1}\left(\frac{x}{a}\right). \]
Using
\[ 1-\sin^2\theta=\cos^2\theta, \]
we obtain
\[ \sqrt{a^2-x^2}=a\cos\theta \]
for $a>0$ and the principal range of $\theta$.
The fraction becomes
\[ \frac{x}{\sqrt{a^2-x^2}} = \tan\theta. \]
Therefore,
\[ \boxed{ \tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right) = \sin^{-1}\left(\frac{x}{a}\right) }. \]
For the expression to be defined,
\[ \boxed{-a < x < a} \]
when $a>0$.
At $x=\pm a$, the denominator becomes zero, so the original expression is not defined.

📘 Concept & Theory

Concept/Theory

›

The numerator and denominator suggest the triple-angle identity for tangent:

\[ \tan3\theta = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}. \]

Since the expression contains both $a^2x$, $x^3$, $a^3$, and $ax^2$, the appropriate substitution is

\[ x=a\tan\theta. \]

Equivalently,

\[ \frac{x}{a}=\tan\theta, \]

so that

\[ \theta=\tan^{-1}\left(\frac{x}{a}\right). \]

This converts the given algebraic expression exactly into the triple-angle form of $\tan3\theta$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Put $x=a\tan\theta$.
Express $x^2$ and $x^3$ in terms of $\tan\theta$.
Substitute these expressions into the numerator and denominator.
Take $a^3$ common from both numerator and denominator.
Recognise the resulting expression as $\tan3\theta$.
Apply $\tan^{-1}$, taking the principal range into consideration.
Use
\[ \theta=\tan^{-1}\left(\frac{x}{a}\right) \]
to obtain the final result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 24 steps

Assume $a\neq0$
Let
\[x=a\tan\theta\]
Dividing by $a$,
\[\frac{x}{a}=\tan\theta\]
Therefore,
\[\theta=\tan^{-1}\left(\frac{x}{a}\right)\]
Now consider the numerator:
\[3a^2x-x^3\]
Substituting $x=a\tan\theta$,
\[\begin{aligned}3a^2x-x^3&=3a^2(a\tan\theta)-(a\tan\theta)^3\\ &=3a^3\tan\theta-a^3\tan^3\theta\\ &=a^3(3\tan\theta-\tan^3\theta)\end{aligned}\]
Now consider the denominator:
\[a^3-3ax^2\]
Substituting $x=a\tan\theta$,
\[\begin{aligned}a^3-3ax^2&=a^3-3a(a\tan\theta)^2\\ &=a^3-3a^3\tan^2\theta\\ &=a^3(1-3\tan^2\theta)\end{aligned}\]
Therefore,
\[\begin{aligned}\frac{3a^2x-x^3}{a^3-3ax^2}&=\frac{a^3(3\tan\theta-\tan^3\theta)}{a^3(1-3\tan^2\theta)}\\ &=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\end{aligned}\]
we obtain
\[ \frac{3a^2x-x^3}{a^3-3ax^2}=\tan3\theta\]
Hence, the given expression becomes
\[\begin{aligned}\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)&=\tan^{-1}(\tan3\theta)\end{aligned}\]
Principal-Value Analysis
The principal range of $\tan^{-1}y$ is
\[-\frac{\pi}{2}<\tan^{-1}y<\frac{\pi}{2}\]
Therefore, the direct relation
\[\tan^{-1}(\tan3\theta)=3\theta\]
is valid only when
\[-\frac{\pi}{2}<3\theta<\frac{\pi}{2}\]
Dividing throughout by $3$,
\[-\frac{\pi}{6}<\theta<\frac{\pi}{6}\]
Since
\[\theta=\tan^{-1}\left(\frac{x}{a}\right)\]
this condition is equivalent to
\[ -\frac{\pi}{6} < tan^{-1}\left(\frac{x}{a}\right) < \frac{\pi}{6}\]
Since $\tan x$ is increasing on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, applying tangent gives
\[-\frac{1}{\sqrt3} < \frac{x}{a} < \frac{1}{\sqrt3}\]
For $a>0$, this becomes
\[-\frac{a}{\sqrt3} < x <\frac{a}{\sqrt3}\]
Thus, within this interval,
\[\begin{aligned}\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)&=3\theta\\ &=3\tan^{-1}\left(\frac{x}{a}\right)\end{aligned}\]
Hence, the simplified result is
\[\boxed{\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)=3\tan^{-1}\left(\frac{x}{a}\right)}\]
for
\[\boxed{a>0,\quad-\frac{a}{\sqrt3} < x <\frac{a}{\sqrt3}}\]
General Principal-Value Form
For arbitrary values of $x$, we cannot always replace $\tan^{-1}(\tan3\theta)$ by $3\theta$. The principal-value result must remain within
\[\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\]
Therefore, the general result can be written as
\[\boxed{\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)=3\tan^{-1}\left(\frac{x}{a}\right)-k\pi}\]
where $k\in\mathbb Z$ is chosen so that the final value lies in the principal range of $\tan^{-1}$

🎯 Exam Significance

Exam Significance

›

This problem tests recognition and application of the tangent triple-angle identity.
The substitution $x=a\tan\theta$ is the central step in the solution.
Careful expansion of both numerator and denominator is necessary to obtain the correct identity.
The principal range of $\tan^{-1}$ must be considered before replacing $\tan^{-1}(\tan3\theta)$ by $3\theta$.
The excluded values of $x$ should be identified because the original expression must be defined.

Significance for Competitive Entrance Exams

The polynomial pattern
\[ 3t-t^3 \]
over
\[ 1-3t^2 \]
is a direct indicator of $\tan3\theta$.
Recognising this pattern can reduce a lengthy algebraic expression to a one-line inverse-trigonometric result.
Principal-value corrections are important in objective questions involving inverse trigonometric functions.
The denominator condition
\[ a^2-3x^2\neq0 \]
provides an immediate domain check.
This problem connects algebraic polynomial forms with multiple-angle trigonometric identities, a recurring technique in entrance examinations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

When the expression contains $a^2x$, $x^3$, $a^3$, and $ax^2$, try
\[ x=a\tan\theta. \]
The correct transformations are
\[ 3a^2x-x^3 = a^3(3\tan\theta-\tan^3\theta) \]
and
\[ a^3-3ax^2 = a^3(1-3\tan^2\theta). \]
Use
\[ \tan3\theta = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}. \]
Therefore,
\[ \tan^{-1}\left( \frac{3a^2x-x^3}{a^3-3ax^2} \right) = \tan^{-1}(\tan3\theta). \]
For
\[ -\frac{\pi}{6}<\theta<\frac{\pi}{6}, \]
the direct result is
\[ \boxed{3\theta}. \]
Since
\[ \theta=\tan^{-1}\left(\frac{x}{a}\right), \]
we obtain
\[ \boxed{ 3\tan^{-1}\left(\frac{x}{a}\right) }. \]
For $a>0$, the direct-result interval is
\[ \boxed{ -\frac{a}{\sqrt3} < x <\frac{a}{\sqrt3} }. \]
The original expression is undefined at
\[ \boxed{x=\pm\frac{a}{\sqrt3}}. \]

📘 Concept & Theory

Concept/Theory

›

This problem involves a nested inverse-trigonometric expression. The most efficient approach is to evaluate the innermost inverse trigonometric function first.

We use the standard principal value

\[ \sin^{-1}\left(\frac12\right)=\frac{\pi}{6}. \]

After finding this value, the expression becomes an ordinary trigonometric expression involving $\cos\frac{\pi}{3}$. Finally, we evaluate the inverse tangent of the resulting number.

The principal range of $\tan^{-1}x$ is

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

Since the final argument is $1$, its principal inverse tangent is $\frac{\pi}{4}$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Evaluate
\[ \sin^{-1}\left(\frac12\right). \]
Multiply the resulting angle by $2$.
Evaluate the cosine of the resulting angle.
Multiply the cosine value by $2$.
Evaluate the inverse tangent of the resulting value.
Use the principal value of $\tan^{-1}1$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

We have
\[\tan^{-1}\left[2\cos\left(2\sin^{-1}\left(\frac12\right)\right)\right]\]
First, evaluate the innermost inverse sine:
\[\sin^{-1}\left(\frac12\right)=\frac{\pi}{6}\]
Therefore,
\[2\sin^{-1}\left(\frac12\right)=2\left(\frac{\pi}{6}\right)\]
Hence,
\[2\sin^{-1}\left(\frac12\right)=\frac{\pi}{3}\]
Substituting this into the original expression,
\[\begin{aligned}\tan^{-1}\left[2\cos\left(2\sin^{-1}\left(\frac12\right)\right)\right]&=\tan^{-1}\left(2\cos\frac{\pi}{3}\right)\end{aligned}\]
Now,
\[\cos\frac{\pi}{3}=\frac12\]
Therefore,
\[\begin{aligned}\tan^{-1}\left(2\cos\frac{\pi}{3}\right)&=\tan^{-1}\left(2\cdot\frac12\right)\\ &=\tan^{-1}(1)\end{aligned}\]
Since
\[\tan\frac{\pi}{4}=1\]
and
\[-\frac{\pi}{2}<\frac{\pi}{4}<\frac{\pi}{2}\]
we have
\[\tan^{-1}(1)=\frac{\pi}{4}\]
Therefore,
\[\boxed{\tan^{-1}\left[2\cos\left(2\sin^{-1}\left(\frac12\right)\right)\right]=\frac{\pi}{4}}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the evaluation of standard inverse-trigonometric values.
It tests the correct order of operations in a nested inverse-trigonometric expression.
The values
\[ \sin^{-1}\frac12=\frac{\pi}{6}, \qquad \cos\frac{\pi}{3}=\frac12, \qquad \tan^{-1}1=\frac{\pi}{4} \]
are fundamental values that should be memorised.
The problem can be solved directly without introducing an unnecessary variable.
Writing every intermediate angle clearly helps avoid errors involving multiplication by $2$.

Significance for Competitive Entrance Exams

This is a quick evaluation problem based on standard inverse-trigonometric values.
Recognising
\[ \sin^{-1}\frac12=\frac{\pi}{6} \]
immediately reduces the complexity of the expression.
Knowing standard-angle values can make such questions solvable within a few seconds.
The problem also tests awareness of principal values of inverse-trigonometric functions.
For objective examinations, evaluating the innermost function first is the safest and fastest approach.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Remember the standard inverse-trigonometric value
\[ \boxed{\sin^{-1}\frac12=\frac{\pi}{6}}. \]
Therefore,
\[ 2\sin^{-1}\frac12=\frac{\pi}{3}. \]
Use
\[ \boxed{\cos\frac{\pi}{3}=\frac12}. \]
Hence,
\[ 2\cos\frac{\pi}{3}=1. \]
Finally,
\[ \boxed{\tan^{-1}1=\frac{\pi}{4}}. \]
The required value is
\[ \boxed{\frac{\pi}{4}}. \]

📘 Concept & Theory

Concept/Theory

›

This problem uses the standard double-angle identities

\[ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta} \]

and

\[ \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}. \]

Therefore, the substitutions

\[ x=\tan\theta \]

and

\[ y=\tan\phi \]

convert the two inverse-trigonometric expressions into angles involving $2\theta$ and $2\phi$.

The outer expression then becomes

\[ \tan(\theta+\phi), \]

which can be evaluated using the tangent addition formula:

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

Important Principal-Value Concept

We must be careful with

\[ \sin^{-1}(\sin2\theta) \]

and

\[ \cos^{-1}(\cos2\phi). \]

These expressions cannot always be replaced directly by $2\theta$ and $2\phi$. The principal ranges are

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

and

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

Thus, for the direct substitutions

\[ \sin^{-1}(\sin2\theta)=2\theta \]

and

\[ \cos^{-1}(\cos2\phi)=2\phi, \]

we require

\[ -\frac{\pi}{2}\leq2\theta\leq\frac{\pi}{2} \]

and

\[ 0\leq2\phi\leq\pi. \]

Equivalently,

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

and

\[ 0\leq\phi\leq\frac{\pi}{2}. \]

Hence, a convenient set of restrictions is

\[ -1\leq x\leq1, \quad y\geq0. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Put $x=\tan\theta$.
Convert $\frac{2x}{1+x^2}$ into $\sin2\theta$.
Use the principal-value condition to obtain
\[ \sin^{-1}(\sin2\theta)=2\theta. \]
Put $y=\tan\phi$.
Convert $\frac{1-y^2}{1+y^2}$ into $\cos2\phi$.
Use the principal-value condition to obtain
\[ \cos^{-1}(\cos2\phi)=2\phi. \]
Substitute both results into the original expression.
Use the tangent addition formula.
Finally, substitute
\[ \tan\theta=x,\qquad \tan\phi=y. \]

✏️ Solution

Complete Solution

›

Step-by-step Solution · 35 steps

Let
\[x=\tan\theta\]
Then
\[\theta=\tan^{-1}x\]
Consider
\[\frac{2x}{1+x^2}\]
Substituting $x=\tan\theta$,
\[\begin{aligned}\frac{2x}{1+x^2}&=\frac{2\tan\theta}{1+\tan^2\theta}\end{aligned}\]
Using
\[1+\tan^2\theta=\sec^2\theta\]
we get
\[\begin{aligned}\frac{2\tan\theta}{1+\tan^2\theta}&=\frac{2\tan\theta}{\sec^2\theta}\\ &=\frac{2\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos^2\theta}}\\ &=2\frac{\sin\theta}{\cos\theta}\cos^2\theta\\ &=2\sin\theta\cos\theta\\ &=\sin2\theta\end{aligned}\]
Therefore,
\[\frac{2x}{1+x^2}=\sin2\theta\]
Hence,
\[\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\sin^{-1}(\sin2\theta)\]
For the direct principal-value result, let
\[-\frac{\pi}{2}\leq2\theta\leq\frac{\pi}{2}\]
Thus,
\[-\frac{\pi}{4}\leq\theta\leq\frac{\pi}{4}\]
Therefore,
\[\sin^{-1}(\sin2\theta)=2\theta\]
Hence,
\[\boxed{\sin^{-1}\left(\frac{2x}{1+x^2}\right)=2\tan^{-1}x}\]
Now let
\[y=\tan\phi\]
Therefore,
\[\phi=\tan^{-1}y\]
Consider
\[\frac{1-y^2}{1+y^2}\]
Substituting $y=\tan\phi$,
\[\begin{aligned}\frac{1-y^2}{1+y^2}&=\frac{1-\tan^2\phi}{1+\tan^2\phi}\end{aligned}\]
Using the double-angle identity
\[\cos2\phi=\frac{1-\tan^2\phi}{1+\tan^2\phi}\]
we obtain
\[\frac{1-y^2}{1+y^2}=\cos2\phi\]
Therefore,
\[\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)=\cos^{-1}(\cos2\phi)\]
For the direct principal-value result, we require
\[0\leq2\phi\leq\pi\]
Therefore,
\[0\leq\phi\leq\frac{\pi}{2}\]
Hence,
\[\cos^{-1}(\cos2\phi)=2\phi\]
Thus,
\[\boxed{\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)=2\tan^{-1}y}\]
Now substitute both results into the given expression:
\[\begin{aligned}&\tan\frac12\left[\sin^{-1}\left(\frac{2x}{1+x^2}\right)+\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right]\\ &=\tan\frac12\left[2\tan^{-1}x+2\tan^{-1}y\right]\end{aligned}\]
Taking $2$ common inside the bracket,
\[\begin{aligned}&=\tan\left[\tan^{-1}x+\tan^{-1}y\right]\end{aligned}\]
Now use
\[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\]
Let
\[A=\tan^{-1}x,\quad B=\tan^{-1}y\]
Then
\[\tan A=x\quad \text{and}\]

\[\tan B=y\]
Therefore,
\[\begin{aligned}\tan\left[\tan^{-1}x+\tan^{-1}y\right]&=\frac{x+y}{1-xy}\end{aligned}\]
Hence, the required value is
\[\boxed{\frac{x+y}{1-xy}}\]
Domain and Principal-Value Conditions
For the expression
\[\frac{2x}{1+x^2}\]
there is no restriction from the denominator because
\[1+x^2>0\]
for every real $x$. However, to use
\[\sin^{-1}(\sin2\theta)=2\theta\]
directly, we require
\[-\frac{\pi}{4}\leq\theta\leq\frac{\pi}{4}\]
Since $x=\tan\theta$, this gives
\[\boxed{-1\leq x\leq1}\]
Similarly, to use
\[\cos^{-1}(\cos2\phi)=2\phi\]
we require
\[0\leq\phi\leq\frac{\pi}{2}\]
Since $y=\tan\phi$, this corresponds to
\[\boxed{y\geq0}\]
Finally, the tangent addition formula requires
\[1-xy\neq0\]
Thus,
\[\boxed{xy\neq1}\]

🎯 Exam Significance

Exam Significance

›

This problem combines double-angle identities with inverse-trigonometric functions.
It tests recognition of
\[ \frac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta \]
and
\[ \frac{1-\tan^2\phi}{1+\tan^2\phi}=\cos2\phi. \]
It tests the correct use of principal ranges for $\sin^{-1}$ and $\cos^{-1}$.
The final step uses the tangent addition formula, making the problem a combination of several important Chapter 2 concepts.
Careful substitution and range analysis are necessary for a complete solution.

Significance for Competitive Entrance Exams

The patterns
\[ \frac{2x}{1+x^2} \]
and
\[ \frac{1-y^2}{1+y^2} \]
should immediately suggest tangent substitutions.
Recognising these double-angle forms significantly reduces calculation time.
The outer factor $\frac12$ is designed to cancel the double angles produced by the inverse functions.
The resulting expression
\[ \tan(\tan^{-1}x+\tan^{-1}y) \]
can be evaluated immediately using the tangent addition formula.
Principal-value restrictions can distinguish a mathematically correct solution from an apparently similar but incomplete one.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Use
\[ x=\tan\theta \]
to obtain
\[ \frac{2x}{1+x^2}=\sin2\theta. \]
Use
\[ y=\tan\phi \]
to obtain
\[ \frac{1-y^2}{1+y^2}=\cos2\phi. \]
Under the appropriate principal-value restrictions,
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x. \]
Similarly,
\[ \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right) = 2\tan^{-1}y. \]
The factor $\frac12$ gives
\[ \tan\left(\tan^{-1}x+\tan^{-1}y\right). \]
Using the tangent addition formula,
\[ \boxed{ \tan\left(\tan^{-1}x+\tan^{-1}y\right) = \frac{x+y}{1-xy} }. \]
For the direct principal-value derivation, a convenient restriction is
\[ \boxed{-1\leq x\leq1,\qquad y\geq0,\qquad xy\neq1}. \]
Most importantly, the intended question must contain
\[ \boxed{\frac{2x}{1+x^2}} \]
rather than
\[ \frac{x}{1+x^2}. \]

📘 Concept & Theory

Concept/Theory

›

The principal range of the inverse sine function is

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

Therefore, the relation

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

is directly valid only when

\[ -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}. \]

Since

\[ \frac{2\pi}{3}>\frac{\pi}{2}, \]

we cannot directly write

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

Instead, we use the identity

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

Thus, $\frac{2\pi}{3}$ is replaced by its supplementary angle

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

which lies within the principal range of $\sin^{-1}$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Check whether $\frac{2\pi}{3}$ lies in the principal range of $\sin^{-1}$.
Since it does not, use
\[ \sin(\pi-\theta)=\sin\theta. \]
Replace $\frac{2\pi}{3}$ by its supplementary angle
\[ \frac{\pi}{3}. \]
Evaluate the inverse sine using its principal range.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

we have
\[\sin^{-1}\left(\sin\frac{2\pi}{3}\right)\]
Since
\[\frac{2\pi}{3}>\frac{\pi}{2}\]
the angle $\frac{2\pi}{3}$ does not belong to the principal range of $\sin^{-1}$
\[-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\]
Therefore, we must first express the sine of b$\frac{2\pi}{3}$ using an angle within the principal range.
Using
\[ \sin(\pi-\theta)=\sin\theta\]
we get
\[\begin{aligned}\sin\frac{2\pi}{3}&=\sin\left(\pi-\frac{2\pi}{3}\right)\\ &=\sin\frac{\pi}{3}\end{aligned}\]
Therefore,
\[\begin{aligned}\sin^{-1}\left(\sin\frac{2\pi}{3}\right)&=\sin^{-1}\left(\sin\frac{\pi}{3}\right)\end{aligned}\]
Since
\[-\frac{\pi}{2}\leq\frac{\pi}{3}\leq\frac{\pi}{2}\]
we can directly apply the inverse sine function:
\[\sin^{-1}\left(\sin\frac{\pi}{3}\right)=\frac{\pi}{3}\]
Hence,
\[ \boxed{\sin^{-1}\left(\sin\frac{2\pi}{3}\right)=\frac{\pi}{3}}\]
Why the Answer Is Not $\frac{2\pi}{3}$
This is a standard principal-value question. Although
\[\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}\]
the inverse sine function must return its value within
\[\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\]
Both $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ have the same sine:
\[\begin{aligned}\sin\frac{\pi}{3}&=\sin\frac{2\pi}{3}\\ &=\frac{\sqrt3}{2}\end{aligned}\]
But only
\[\frac{\pi}{3}\]
lies within the principal range of $\sin^{-1}$. Therefore, the inverse sine selects $\frac{\pi}{3}$

🎯 Exam Significance

Exam Significance

›

This problem directly tests the concept of principal values of inverse trigonometric functions.
It is important to check the range before simplifying expressions of the form $\sin^{-1}(\sin\theta)$.
The identity
\[ \sin(\pi-\theta)=\sin\theta \]
is the key step.
This is a short question, but a range-based explanation makes the solution mathematically complete.

Significance for Competitive Entrance Exams

Principal-value questions are common conceptual traps in entrance examinations.
The fastest method is to compare the given angle with the principal range of the inverse function.
For an angle in the second quadrant, the corresponding principal inverse-sine value is its supplementary angle.
Memorising the principal ranges of $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ is essential for avoiding incorrect answers.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The principal range of $\sin^{-1}x$ is
\[ \boxed{ -\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2} }. \]
Since
\[ \frac{2\pi}{3}>\frac{\pi}{2}, \]
we cannot directly use
\[ \sin^{-1}(\sin\theta)=\theta. \]
Use
\[ \sin(\pi-\theta)=\sin\theta. \]
Thus,
\[ \sin\frac{2\pi}{3} = \sin\frac{\pi}{3}. \]
Since
\[ \frac{\pi}{3} \]
lies within the principal range of $\sin^{-1}$,
\[ \sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3}. \]
Therefore,
\[ \boxed{ \sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3} }. \]

📘 Concept & Theory

Concept/Theory

›

This problem tests the concept of the principal value of an inverse trigonometric function.

The principal range of the inverse tangent function is

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

Therefore, the relation

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

is directly valid only when

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

Here,

\[ \theta=\frac{3\pi}{4}, \]

and

\[ \frac{3\pi}{4}>\frac{\pi}{2}. \]

Hence, we cannot directly write

\[ \tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \frac{3\pi}{4}. \]

We must find the coterminal angle having the same tangent that lies within the principal range of $\tan^{-1}$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the principal range of $\tan^{-1}x$.
Check whether $\frac{3\pi}{4}$ lies in this range.
Since it does not, find an equivalent angle with the same tangent that lies in the principal range.
Use
\[ \tan(\theta-\pi)=\tan\theta. \]
Evaluate the resulting principal angle.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

We have
\[\tan^{-1}\left(\tan\frac{3\pi}{4}\right)\]
First, observe that
\[\frac{3\pi}{4}>\frac{\pi}{2}\]
Therefore, $\frac{3\pi}{4}$ does not lie in the principal range of $\tan^{-1}$
The principal range of $\tan^{-1}x$ is
\[-\frac{\pi}{2}<\tan^{-1}x<\frac{\pi}{2}\]
Now use the periodicity of the tangent function:
\[\tan(\theta-\pi)=\tan\theta\]
Therefore,
\[\begin{aligned}\tan\frac{3\pi}{4}&=\tan\left(\frac{3\pi}{4}-\pi\right)\\ &=\tan\left(-\frac{\pi}{4}\right)\end{aligned}\]
Hence,
\[\begin{aligned}\tan^{-1}\left(\tan\frac{3\pi}{4}\right)&=\tan^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right)\end{aligned}\]
Since
\[-\frac{\pi}{2} < -\frac{\pi}{4} < \frac{\pi}{2}\]
the angle $-\frac{\pi}{4}$ lies within the principal range of $\tan^{-1}$
Therefore,
\[ \tan^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right)=-\frac{\pi}{4}\]
Hence, the required value is
\[\boxed{\tan^{-1}\left(\tan\frac{3\pi}{4}\right)=-\frac{\pi}{4}}\]
General Rule
For any real $\theta$
\[\tan^{-1}(\tan\theta)\]
is the unique angle in
\[\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\]
that has the same tangent as $\theta$
In particular,
\[ \boxed{\tan^{-1}(\tan\theta)=\theta-k\pi}\]
where $k\in\mathbb Z$ is chosen so that
\[-\frac{\pi}{2}<\theta-k\pi<\frac{\pi}{2}\]

🎯 Exam Significance

Exam Significance

›

This is a direct test of the principal range of $\tan^{-1}x$.
Students should never automatically cancel $\tan^{-1}$ and $\tan$.
The principal range
\[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]
must be remembered.
The problem is short but conceptually important because an incorrect range assumption changes the final answer.
Writing the range explicitly makes the reasoning complete and exam-ready.

Significance for Competitive Entrance Exams

Principal-value traps are common in objective questions on inverse trigonometric functions.
The quickest method is to reduce the angle to the principal range while preserving its tangent.
Since
\[ \tan(\theta-\pi)=\tan\theta, \]
subtracting $\pi$ from $\frac{3\pi}{4}$ immediately gives the required principal angle.
Recognising
\[ \tan\frac{3\pi}{4}=-1 \]
provides an even faster verification.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The principal range of $\tan^{-1}x$ is
\[ \boxed{ -\frac{\pi}{2}<\tan^{-1}x<\frac{\pi}{2} }. \]
The angle
\[ \frac{3\pi}{4} \]
is outside this principal range.
Use the periodicity of tangent:
\[ \tan(\theta-\pi)=\tan\theta. \]
Therefore,
\[ \tan\frac{3\pi}{4} = \tan\left(-\frac{\pi}{4}\right). \]
Since
\[ -\frac{\pi}{4} \]
lies within the principal range of $\tan^{-1}$,
\[ \tan^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right) = -\frac{\pi}{4}. \]
Thus,
\[ \boxed{ \tan^{-1}\left(\tan\frac{3\pi}{4}\right) = -\frac{\pi}{4} }. \]

📘 Concept & Theory

Concept/Theory

›

This problem combines inverse trigonometric functions with the tangent addition formula.

The principal range of $\sin^{-1}x$ is

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

while the principal range commonly used for $\cot^{-1}x$ is

\[ 0<\cot^{-1}x<\pi. \]

Since both $\frac35$ and $\frac32$ are positive, the corresponding angles lie in the first quadrant. This allows us to construct right-triangle ratios directly.

We then use

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Let
\[ \theta_1=\sin^{-1}\left(\frac35\right). \]
Use a $3$-$4$-$5$ right triangle to find
\[ \tan\theta_1=\frac34. \]
Let
\[ \theta_2=\cot^{-1}\left(\frac32\right). \]
Use the definition of cotangent to obtain
\[ \tan\theta_2=\frac23. \]
Apply the tangent addition formula.
Simplify the resulting fraction.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Let
\[\theta_1=\sin^{-1}\left(\frac35\right)\]
Therefore,
\[\sin\theta_1=\frac35\]
Since $\theta_1$ lies in the first quadrant, consider a right triangle with
\[\text{opposite}=3,\quad\text{hypotenuse}=5\]
By the Pythagorean theorem, the adjacent side is
\[\begin{aligned}\text{adjacent}&=\sqrt{5^2-3^2}\\ &=\sqrt{25-9}\\ =\sqrt{16}\\ =4\end{aligned}\]
Therefore,
\[\begin{aligned}\tan\theta_1&=\frac{\text{opposite}}{\text{adjacent}}\\ &=\frac34\end{aligned}\]
Hence,
\[\boxed{\tan\theta_1=\frac34}\]
Now let
\[\theta_2=\cot^{-1}\left(\frac32\right)\]
Therefore,
\[\cot\theta_2=\frac32\]
Using
\[\cot\theta_2=\frac{1}{\tan\theta_2}\]
we obtain
\[\tan\theta_2=\frac{1}{\cot\theta_2}\]
Thus,
\[\begin{aligned}\tan\theta_2&=\frac{1}{\frac32}\\&=\frac23\end{aligned}\]
Hence,
\[\boxed{\tan\theta_2=\frac23}\]
The required expression is
\[\tan(\theta_1+\theta_2)\]
Using the tangent addition formula
\[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\]
we get
\[\begin{aligned}\tan(\theta_1+\theta_2)&=\frac{\tan\theta_1+\tan\theta_2}{1-\tan\theta_1\tan\theta_2}\end{aligned}\]
Substituting
\[\tan\theta_1=\frac34,\quad\tan\theta_2=\frac23\]
we obtain
\[\begin{aligned}\tan(\theta_1+\theta_2)&=\frac{\frac34+\frac23}{1-\frac34\cdot\frac23}\end{aligned}\]
Taking the numerator first:
\[\begin{aligned}\frac34+\frac23&=\frac{9+8}{12}\\&=\frac{17}{12}\end{aligned}\]
Now simplify the denominator:
\[\begin{aligned}1-\frac34\cdot\frac23&=1-\frac{6}{12}\\ &=1-\frac12\\ &=\frac12\end{aligned}\]
Therefore,
\[\begin{aligned}\tan(\theta_1+\theta_2)&=\frac{\frac{17}{12}}{\frac12}\\ &=\frac{17}{12}\times2\\ &=\frac{17}{6}\end{aligned}\]
Hence, the required value is
\[\boxed{\frac{17}{6}}\]

🎯 Exam Significance

Exam Significance

›

This problem tests conversion between inverse trigonometric functions and ordinary trigonometric ratios.
The $3$-$4$-$5$ triangle is a standard and useful shortcut for
\[ \sin\theta=\frac35. \]
The reciprocal relationship
\[ \cot\theta=\frac{1}{\tan\theta} \]
is essential for handling the second inverse function.
The final step tests the tangent addition formula.
Writing the intermediate values of both tangent ratios clearly makes the solution easy to verify and reduces calculation errors.

Significance for Competitive Entrance Exams

The expression can be reduced rapidly by recognising the $3$-$4$-$5$ triangle.
For
\[ \cot^{-1}\left(\frac32\right), \]
immediately use
\[ \tan\theta=\frac23. \]
The tangent addition formula then gives the answer in a few steps.
Checking the denominator
\[ 1-\frac34\cdot\frac23=\frac12 \]
confirms that the addition formula is valid.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

If
\[ \theta_1=\sin^{-1}\left(\frac35\right), \]
then
\[ \boxed{\tan\theta_1=\frac34}. \]
If
\[ \theta_2=\cot^{-1}\left(\frac32\right), \]
then
\[ \boxed{\tan\theta_2=\frac23}. \]
Use
\[ \tan(A+B) = \frac{\tan A+\tan B} {1-\tan A\tan B}. \]
Therefore,
\[ \tan(\theta_1+\theta_2) = \frac{\frac34+\frac23} {1-\frac34\cdot\frac23}. \]
Simplifying,
\[ \frac{\frac{17}{12}}{\frac12} = \frac{17}{6}. \]
Hence, the required value is
\[ \boxed{\frac{17}{6}}. \]

📘 Concept & Theory

Concept/Theory

›

This problem tests the principal value of the inverse cosine function.

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

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

Therefore, the relation

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

is directly valid only when

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

Here,

\[ \frac{7\pi}{6}>\pi. \]

Hence, we cannot directly write

\[ \cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{7\pi}{6}. \]

We must find an angle in the principal range $[0,\pi]$ having the same cosine.

Since cosine satisfies

\[ \cos(2\pi-\theta)=\cos\theta, \]

we can replace $\frac{7\pi}{6}$ by

\[ 2\pi-\frac{7\pi}{6} = \frac{5\pi}{6}. \]

The angle $\frac{5\pi}{6}$ lies within the principal range of $\cos^{-1}$, so it is the required principal value.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the principal range of $\cos^{-1}x$.
Check whether $\frac{7\pi}{6}$ lies within that range.
Since it does not, find an angle in $[0,\pi]$ having the same cosine.
Use
\[ \cos(2\pi-\theta)=\cos\theta. \]
Evaluate the resulting principal angle.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

We have
\[\cos^{-1}\left(\cos\frac{7\pi}{6}\right)\]
The principal range of $\cos^{-1}x$ is
\[0\leq\cos^{-1}x\leq\pi\]
However,
\[\frac{7\pi}{6}>\pi\]
Therefore, $\frac{7\pi}{6}$ is outside the principal range, and we cannot directly use
\[\cos^{-1}(\cos\theta)=\theta\]
Using the identity
\[\cos(2\pi-\theta)=\cos\theta\]
we get
\[\begin{aligned}\cos\frac{7\pi}{6}&=\cos\left(2\pi-\frac{7\pi}{6}\right)\\ &=\cos\left(\frac{12\pi-7\pi}{6}\right)\\ &=\cos\frac{5\pi}{6}\end{aligned}\]
Therefore,
\[\begin{aligned}\cos^{-1}\left(\cos\frac{7\pi}{6}\right)&=\cos^{-1}\left(\cos\frac{5\pi}{6}\right)\end{aligned}\]
Now,
\[0\leq\frac{5\pi}{6}\leq\pi\]
Thus, $\frac{5\pi}{6}$ lies within the principal range of $\cos^{-1}$. Hence,
\[\cos^{-1}\left(\cos\frac{5\pi}{6}\right)=\frac{5\pi}{6}\]
Therefore, the required value is
\[\boxed{\cos^{-1}\left(\cos\frac{7\pi}{6}\right)=\frac{5\pi}{6}}\]

🎯 Exam Significance

Exam Significance

›

This problem directly tests the principal range of $\cos^{-1}x$.
Students must not automatically cancel $\cos^{-1}$ and $\cos$.
The principal range
\[ [0,\pi] \]
must be checked before applying
\[ \cos^{-1}(\cos\theta)=\theta. \]
The identity
\[ \cos(2\pi-\theta)=\cos\theta \]
provides a convenient way to bring the angle into the required range.

Significance for Competitive Entrance Exams

Principal-value questions are common conceptual traps in inverse trigonometry.
The fastest approach is to identify an equivalent angle within the principal range.
For
\[ \frac{7\pi}{6}, \]
subtracting it from $2\pi$ immediately gives
\[ \frac{5\pi}{6}. \]
The result can also be verified by recognising
\[ \cos\frac{7\pi}{6} = -\frac{\sqrt3}{2}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The principal range of $\cos^{-1}x$ is
\[ \boxed{ 0\leq\cos^{-1}x\leq\pi }. \]
The angle
\[ \frac{7\pi}{6} \]
lies outside this principal range.
Use
\[ \cos(2\pi-\theta)=\cos\theta. \]
Therefore,
\[ \cos\frac{7\pi}{6} = \cos\frac{5\pi}{6}. \]
Since
\[ \frac{5\pi}{6}\in[0,\pi], \]
we have
\[ \cos^{-1}\left(\cos\frac{5\pi}{6}\right) = \frac{5\pi}{6}. \]
Hence,
\[ \boxed{ \cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6} }. \]

📘 Concept & Theory

Concept/Theory

›

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

The principal range of $\sin^{-1}x$ is

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

Since

\[ \sin\left(-\frac{\pi}{6}\right)=-\frac12 \]

and

\[ -\frac{\pi}{2}\leq-\frac{\pi}{6}\leq\frac{\pi}{2}, \]

we have

\[ \sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6}. \]

The expression then reduces to a standard sine value.

🗺️ Solution Roadmap

Step-by-step Plan

›

Evaluate
\[ \sin^{-1}\left(-\frac12\right). \]
Substitute its principal value into the given expression.
Convert the subtraction of a negative angle into addition.
Simplify the resulting angle.
Evaluate the standard sine value.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

We have
\[\sin\left[\frac{\pi}{3}-\sin^{-1}\left(-\frac12\right)\right]\]
First, evaluate the inverse sine:
\[\sin^{-1}\left(-\frac12\right)\]
Since
\[ \sin\left(-\frac{\pi}{6}\right)=-\frac12\]
and $-\frac{\pi}{6}$ lies within the principal range of $\sin^{-1}$,
\[\boxed{\sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}}\]
Substituting this value,
\[\begin{aligned}\sin\left[\frac{\pi}{3}-\sin^{-1}\left(-\frac12\right)\right]&=\sin\left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right]\end{aligned}\]
Subtracting a negative number is equivalent to addition:
\[\begin{aligned}&=\sin\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\end{aligned}\]
Taking the LCM of the denominators,
\[\begin{aligned}\frac{\pi}{3}+\frac{\pi}{6}&=\frac{2\pi}{6}+\frac{\pi}{6}\\&=\frac{3\pi}{6}\\&=\frac{\pi}{2}\end{aligned}\]
Therefore,
\[\begin{aligned}\sin\left(\frac{\pi}{3}+\frac{\pi}{6}\right)&=\sin\frac{\pi}{2}\\&=1\end{aligned}\]
Hence, the required value is
\[\boxed{1}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the principal value of the inverse sine function.
It reinforces the standard value
\[ \sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}. \]
Students must carefully handle the subtraction of a negative angle.
The final step uses the standard value
\[ \sin\frac{\pi}{2}=1. \]
Writing the intermediate angle simplification prevents sign and fraction errors.

Significance for Competitive Entrance Exams

The problem can be solved quickly by recognising the standard inverse-sine value.
The negative sign inside $\sin^{-1}$ is a common source of errors, so the principal value should be written explicitly.
The transformation
\[ \frac{\pi}{3}-\left(-\frac{\pi}{6}\right) = \frac{\pi}{2} \]
immediately identifies the final standard value.
Such questions test conceptual accuracy rather than lengthy calculation.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The principal value is
\[ \boxed{ \sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6} }. \]
Subtracting a negative angle gives
\[ \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3}+\frac{\pi}{6}. \]
Therefore,
\[ \frac{\pi}{3}+\frac{\pi}{6} = \frac{\pi}{2}. \]
Using
\[ \sin\frac{\pi}{2}=1, \]
we obtain the required value.
Hence,
\[ \boxed{ \sin\left[ \frac{\pi}{3} - \sin^{-1}\left(-\frac12\right) \right] = 1 } \]

📘 Concept & Theory

Concept/Theory

›

This problem tests the principal values of $\tan^{-1}x$ and $\cot^{-1}x$.

The principal range of $\tan^{-1}x$ is

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

Therefore,

\[ \tan^{-1}(\sqrt3)=\frac{\pi}{3}, \]

because

\[ \tan\frac{\pi}{3}=\sqrt3. \]

For the inverse cotangent function, using the NCERT convention, its principal range is

\[ 0<\cot^{-1}x<\pi. \]

We need an angle in this interval whose cotangent is $-\sqrt3$.

Since

\[ \cot\frac{5\pi}{6} = -\sqrt3, \]

we have

\[ \cot^{-1}(-\sqrt3)=\frac{5\pi}{6}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Evaluate
\[ \tan^{-1}(\sqrt3). \]
Evaluate
\[ \cot^{-1}(-\sqrt3) \]
using its principal range.
Substitute both values into the given expression.
Simplify the resulting angle.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

We have
\[\tan^{-1}(\sqrt3)-\cot^{-1}(-\sqrt3)\]
First, consider
\[\tan^{-1}(\sqrt3)\]
Since
\[\tan\frac{\pi}{3}=\sqrt3\quad\text{and}\]

\[-\frac{\pi}{2}<\frac{\pi}{3}<\frac{\pi}{2}\]
$\frac{\pi}{3}$ lies within the principal range of $\tan^{-1}$. Therefore,
\[\boxed{\tan^{-1}(\sqrt3)=\frac{\pi}{3}}\]
Now consider
\[\cot^{-1}(-\sqrt3)\]
The principal range of $\cot^{-1}x$ is
\[0<\cot^{-1}x<\pi\]
We know that
\[\tan\frac{\pi}{6}=\frac{1}{\sqrt3}\]
Therefore,
\[\cot\frac{\pi}{6}=\sqrt3\]
Since cotangent is negative in the second quadrant, the required principal angle is
\[\pi-\frac{\pi}{6}=\frac{5\pi}{6}\]
Hence,
\[\boxed{\cot^{-1}(-\sqrt3)=\frac{5\pi}{6}}\]
Substituting both values into the original expression:
\[\begin{aligned}\tan^{-1}(\sqrt3)-\cot^{-1}(-\sqrt3)&=\frac{\pi}{3}-\frac{5\pi}{6}\end{aligned}\]
Expressing $\frac{\pi}{3}$ with denominator $6$:
\[\begin{aligned}\frac{\pi}{3} - \frac{5\pi}{6}&=\frac{2\pi}{6}-\frac{5\pi}{6}\\ &=-\frac{3\pi}{6}\\ &=-\frac{\pi}{2}\end{aligned}\]
herefore, the required value is
\[\boxed{-\frac{\pi}{2}}\]

🎯 Exam Significance

Exam Significance

›

This question is primarily based on the principal values of inverse trigonometric functions.
The most important point is the principal range of $\cot^{-1}x$.
A negative value of cotangent corresponds to an angle in the second quadrant when the principal range is $(0,\pi)$.
Students should not automatically use a negative acute angle for $\cot^{-1}$ of a negative number.
Remembering the principal ranges prevents sign errors in otherwise simple questions.

Significance for Competitive Entrance Exams

This is a classic principal-value trap in inverse trigonometry.
The quickest reliable method is to determine the quadrant from the sign of the trigonometric ratio and then select the angle within the principal range.
For
\[ \cot^{-1}(-\sqrt3), \]
the reference angle is
\[ \frac{\pi}{6}, \]
and the required principal angle is
\[ \pi-\frac{\pi}{6}=\frac{5\pi}{6}. \]
Careful treatment of principal values is essential in JEE and other entrance examinations.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 5 points

\[ \boxed{ \tan^{-1}(\sqrt3)=\frac{\pi}{3} }. \]
Under the standard NCERT convention,
\[ \boxed{ 0<\cot^{-1}x<\pi }. \]
Since
\[ \cot\frac{5\pi}{6}=-\sqrt3, \]
we have
\[ \boxed{ \cot^{-1}(-\sqrt3)=\frac{5\pi}{6} }. \]
Therefore,
\[ \begin{aligned} \tan^{-1}(\sqrt3) - \cot^{-1}(-\sqrt3) &= \frac{\pi}{3} - \frac{5\pi}{6}\ &= -\frac{\pi}{2}. \end{aligned} \]
Hence, the correct answer is
\[ \boxed{ -\frac{\pi}{2} }. \]

Inverse Trigonometric Functions

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Domain of the Given Expression

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Important Principal-Value Concept

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Chapter Complete!

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Relations and Functions — Learning Resources

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Inverse Trigonometric Functions

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Domain of the Given Expression

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Important Principal-Value Concept

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Chapter Complete!

Recent posts

Relations and Functions — Learning Resources

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