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
Solution
<p>$$\begin{aligned}
& \frac{\pi}{2} \leq x \leq \frac{3 \pi}{4} \\
& \cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{12} \sin x\right) \\
& \cos ^{-1}(\cos x \cos \alpha+\sin x \sin \alpha) \\
& \cos ^{-1}(\cos (x-\alpha)) \\
& \Rightarrow x-\alpha \text { because } x-\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\
& \Rightarrow x-\tan ^{-1} \frac{5}{12}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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