JEE MAIN - Mathematics (2025 - 23rd January Morning Shift - No. 6)
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
$x+\tan ^{-1} \frac{5}{12}$
$x-\tan ^{-1} \frac{4}{3}$
$x+\tan ^{-1} \frac{4}{5}$
$x-\tan ^{-1} \frac{5}{12}$
Explanation
$$\begin{aligned}
& \frac{\pi}{2} \leq x \leq \frac{3 \pi}{4} \\
& \cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{12} \sin x\right) \\
& \cos ^{-1}(\cos x \cos \alpha+\sin x \sin \alpha) \\
& \cos ^{-1}(\cos (x-\alpha)) \\
& \Rightarrow x-\alpha \text { because } x-\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\
& \Rightarrow x-\tan ^{-1} \frac{5}{12}
\end{aligned}$$
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