Question

If $x\in (-\frac{\pi }{2},\frac{3\pi }{2}),$ then ${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ is equal to

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

Answer 1

The correct option is B $\frac{\pi }{4}-\frac{x}{2}$

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

As $x\in (-\frac{\pi }{2},\frac{3\pi }{2}),$
$\frac{\pi }{4}-\frac{x}{2}\in (-\frac{\pi }{2},\frac{\pi }{2})$
$\therefore {\tan }^{-1}\tan (\frac{\pi }{4}-\frac{x}{2})=\frac{\pi }{4}-\frac{x}{2}$