Question ID: #242
If $\frac{\pi}{2} \le x \le \frac{3\pi}{4}$, then $\cos^{-1}\left(\frac{12}{13}\cos x + \frac{5}{13}\sin x\right)$ is equal to
- (1) $x – \tan^{-1}\frac{4}{3}$
- (2) $x – \tan^{-1}\frac{5}{12}$
- (3) $x + \tan^{-1}\frac{4}{5}$
- (4) $x + \tan^{-1}\frac{5}{12}$
Solution:
Let $\cos \alpha = \frac{12}{13}$ and $\sin \alpha = \frac{5}{13}$.
Then $\tan \alpha = \frac{5}{12} \implies \alpha = \tan^{-1}\frac{5}{12}$.
Expression becomes:
$y = \cos^{-1}(\cos x \cos \alpha + \sin x \sin \alpha)$
$y = \cos^{-1}(\cos(x – \alpha))$
$y = x – \alpha$.
$y = x – \tan^{-1}\frac{5}{12}$.
Ans. (2)
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