Question

Prove that ${\tan }^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)=\frac{\pi }{4}-\frac{x}{2}$ if $\pi <x<\frac{3\pi }{2}$

Answer 1

${\tan }^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)$

$={\tan }^{-1}\left(\frac{{(\sqrt{1+\cos x}+\sqrt{1-\cos x})}^2}{{\left(\sqrt{1+\cos x}\right)}^2-{\left(\sqrt{1-\cos x}\right)}^2}\right)$

$={\tan }^{-1}\left(\frac{1+\cos x+1-\cos x+2\sqrt{1-\cos {}^{2}x}}{1+\cos x-1+\cos x}\right)$

$={\tan }^{-1}\left(\frac{2+2\sqrt{1-\sin {}^{2}x}}{2\cos x}\right)$

$={\tan }^{-1}\left(\frac{1+\sin x}{\cos x}\right)$

$={\tan }^{-1}\left(\frac{1+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}\right)$

$={\tan }^{-1}\left(\frac{{(1+\tan \frac{x}{2})}^2(1+\tan \frac{x}{2})}{{(1-\tan \frac{x}{2})}^2(1+\tan \frac{x}{2})}\right)$

$={\tan }^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)$

$={\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}+\tan \frac{x}{2}}{1-\tan \frac{x}{4}\tan \frac{x}{2}}\right)$

As $\pi <x<\frac{3x}{2}$

$={\tan }^{-1}\left(\tan (\frac{\pi }{4}-\frac{x}{2})\right)$

$=\frac{\pi }{4}-\frac{x}{2}$

Hence, the answer is $\frac{\pi }{4}-\frac{x}{2}.$