Find the simplified form of $\cos^{-1} \left( \frac{3}{5} \cos x + \frac{4}{5} \sin x \right)$, where $x \in \left[ \frac{-3\pi}{4}, \frac{\pi}{4} \right]$.

$\tan^{-1}\left(\frac{4}{3}\right) - x$

The correct answer is Option (2) → $\tan^{-1}\left(\frac{4}{3}\right) - x$ ##

We have, $\cos^{-1} \left[ \frac{3}{5} \cos x + \frac{4}{5} \sin x \right], x \in \left[ \frac{-3\pi}{4}, \frac{\pi}{4} \right]$

Let $\cos y = \frac{3}{5}$

$\Rightarrow \sin y = \frac{4}{5}$

$\Rightarrow y = \cos^{-1} \frac{3}{5} = \sin^{-1} \frac{4}{5} = \tan^{-1} \left( \frac{4}{3} \right)$

$∴\cos^{-1} \left( \frac{3}{5} \cos x + \frac{4}{5} \sin x \right) = \cos^{-1} [\cos y \cdot \cos x + \sin y \cdot \sin x]$

$= \cos^{-1} [\cos (y - x)] \quad [∵\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B]$

$= y - x = \tan^{-1} \frac{4}{3} - x \quad \left[ ∵y = \tan^{-1} \frac{4}{3} \right]$