Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Assume that ,
$x^2=\cos 2\theta ;\theta =\frac{1}{2}{\cos }^{-1}{x}^2$

$$\begin{gathered} \therefore {\tan }^{-\; 1}\left[\frac{\sqrt{1+\cos 2\theta }+ \sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }- \sqrt{1-\cos 2\theta }}\right] \\ ={\tan }^{-\; 1}\left[\frac{\sqrt{2{\cos }^2\theta }+ \sqrt{2{\sin }^2\theta }}{\sqrt{2{\cos }^2\theta }- \sqrt{2{\sin }^2\theta }}\right] \\ ={\tan }^{-\; 1}\left[\frac{\cos \theta + \sin \theta }{{\cos }^{}\theta - {\sin }^{}\theta }\right] \end{gathered}$$
$={\tan }^{-1}\left[\frac{1+\tan \theta }{1-\tan \theta }\right]$

$={\tan }^{-1}\left[\tan \left(\frac{\pi }{4}+\theta \right)\right]$

$=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2$