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

(B) $x-\tan ^{-1} \frac{5}{12}$

(D) $x+\tan ^{-1} \frac{5}{12}$

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Inverse Trigonometric Functions

\frac{π}{2} \leq x \leq \frac{3 π}{4}

, then

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

is equal to

Answer: (B)

Write

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

, where

cos α = \frac{12}{13}

sin α = \frac{5}{13}

, so

α = tan^{- 1} \frac{5}{12}

.

For

\frac{π}{2} \leq x \leq \frac{3 π}{4}

and

α = tan^{- 1} \frac{5}{12} \in (0, \frac{π}{2})

x - α \in (\frac{π}{2} - α, \frac{3 π}{4} - α) \subset [0, π]

Since

x - α

lies in

[0, π]

, the principal range of

cos^{- 1}

cos^{- 1} (cos (x - α)) = x - α = x - tan^{- 1} \frac{5}{12}

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