Question

Simplify:

${\tan }^{-1}\left(\frac{1-\cos x}{\sin x}\right)$ for $x\in (0,\frac{\pi }{2})$

• A. $\frac{x}{4}$
• B. $x$
• C. $2x$
• D. $\frac{x}{2}$

Answer 1

Answer: (D)

$\frac{x}{2}$

Consider given the expression, ${\tan }^{-1}\left(\frac{1-\cos x}{\sin x}\right),x\in (0,\frac{\pi }{2})$

$\because \cos x=1-2{\sin }^2\frac{x}{2},\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

Therefore,

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

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

$=\frac{x}{2}$

Hence, this is the required result.