Question

Write the simplest form of ${\tan }^{-1}\left(\begin{array}{c}\sqrt{\frac{1-\cos x}{1+\cos x}}\end{array}\right)0<x<\pi .$.

Answer 1

Given that $0<x<\pi$

We have to simplify ${\tan }^{-1}\left(\sqrt{\frac{1-cosx}{1+cosx}}\right)$

We know that $\cos x=2{\cos }^2\left(\frac{x}{2}\right)-1=1-2{\sin }^2\left(\frac{x}{2}\right)$

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

Given that $0<x<\pi$ , which implies $0<\frac{x}{2}<\frac{\pi }{2}$

So we get ${\tan }^{-1}\left(\sqrt{\frac{1-cosx}{1+cosx}}\right)={\tan }^{-1}\left(\sqrt{{\tan }^2\left(\frac{x}{2}\right)}\right)={\tan }^{-1}\left(\tan \left(\frac{x}{2}\right)\right)=\frac{x}{2}$

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