If 2 x 3 4, then ^-1 ( 1213 x + 513 x) is equal to: — Solution | JEE Mathematics

MCQMediumJEE 2025Inverse Trigonometric Functions

If $\frac{\pi}{2} \leq x \leq \frac{3\pi}{4}$ , then $\cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right)$ is equal to:

Answer

Correct answer:B

Standard Method

Given: $\frac{\pi}{2} \leq x \leq \frac{3\pi}{4}$ and we need to evaluate $\cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right)$ .

Find: Which given option equals the expression.

Let

a = \frac{12}{13}, \qquad b = \frac{5}{13}

Then

\sqrt{a^2+b^2} = \sqrt{\left(\frac{12}{13}\right)^2 + \left(\frac{5}{13}\right)^2} = 1

So we can write

\frac{12}{13} \cos x + \frac{5}{13} \sin x = \cos\theta \cos x + \sin\theta \sin x = \cos(x-\theta)

where

\theta = \tan^{-1}\left(\frac{5}{12}\right)

Hence the given expression becomes

\cos^{-1}(\cos(x-\theta))

Since the solution states that the angle lies in the range where the principal value gives

\cos^{-1}(\cos(x-\theta)) = x-\theta

we get

\cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right) = x - \tan^{-1}\left(\frac{5}{12}\right)

Therefore, the correct option is B.

Identity Recognition

Given: $\cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right)$ with $\frac{\pi}{2} \leq x \leq \frac{3\pi}{4}$ .

Find: The equivalent simplified form.

Observe that

\cos\theta = \frac{12}{13}, \qquad \sin\theta = \frac{5}{13}

for

\theta = \tan^{-1}\left(\frac{5}{12}\right)

Using

\cos(A-B) = \cos A \cos B + \sin A \sin B

we obtain

\frac{12}{13} \cos x + \frac{5}{13} \sin x = \cos(x-\theta)

Therefore,

\cos^{-1} \left( \frac{12}{13} \cos x + \frac{5}{13} \sin x \right) = \cos^{-1}(\cos(x-\theta))

and from the given range used in the source solution,

\cos^{-1}(\cos(x-\theta)) = x-\theta

So the expression equals $x - \tan^{-1}\left(\frac{5}{12}\right)$ , which matches option B.

Taking $\frac{12}{13} \cos x + \frac{5}{13} \sin x$ as $\cos(x+\theta)$ . This is wrong because $\cos(x+\theta)=\cos x\cos\theta-\sin x\sin\theta$ has a minus sign before the sine term. Use $\cos(x-\theta)=\cos x\cos\theta+\sin x\sin\theta$ instead.
Choosing $\theta = \tan^{-1}\left(\frac{12}{5}\right)$ instead of $\tan^{-1}\left(\frac{5}{12}\right)$ . This reverses the roles of sine and cosine. Since $\tan\theta=\frac{\sin\theta}{\cos\theta}$ , here it must be $\frac{5/13}{12/13}=\frac{5}{12}$ .
Using $\cos^{-1}(\cos t)=t$ without checking the principal value range. The inverse cosine returns values in $[0,\pi]$ , so the angle must be interpreted with its valid range. Here the given interval ensures the source solution’s step is valid.

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