Question

The derivative of f(x) defined by $f\left(x\right)={\tan }^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\pi <x<\pi$
is

• A. $1,\mathrm{if}0<x<\pi$
• B. $-1,\mathrm{if}-\pi <x<0$
• C. $\frac{1}{2},\mathrm{if}0<x<\pi$
• D. $-\frac{1}{2},\mathrm{if}-\pi <x<0$

Answer 1

Answer: (C), (D)

The correct options are
C $\frac{1}{2},\mathrm{if}0<x<\pi$
D $-\frac{1}{2},\mathrm{if}-\pi <x<0$

Given that,
$\mathrm{Let}y={\tan }^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\pi <x<\pi \mathrm{Then},y={\tan }^{-1}\left(\sqrt{\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}}\right)={\tan }^{-1}\left(\tan \frac{x}{2}\right|\Rightarrow y=(\begin{array}{c}{\tan }^{-1}\left(\tan \frac{x}{2}\right),\mathrm{if}\tan \frac{x}{2}\ge 0\\ {\tan }^{-1}\left(-\tan \frac{x}{2}\right),\mathrm{if}\tan \frac{x}{2}<0\end{array}\Rightarrow y=(\begin{array}{c}\frac{x}{2},\mathrm{if}0\le x<\pi \\ -\frac{x}{2},\mathrm{if}-\pi <x<0\end{array}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=(\begin{array}{c}\frac{1}{2},\mathrm{if}0\le x<\pi \\ -\frac{1}{2},\mathrm{if}-\pi <x<0\end{array}$