Question

Show that ${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\frac{\pi }{p}-\frac{x}{q},-\frac{\pi }{2}<x<\frac{3\pi }{2}$.
Find the value of p and q

• A. $p=2,q=3$
• B. $p=3,q=2$
• C. $p=2,q=4$
• D. $p=4,q=2$

Answer 1

The correct option is D $p=4,q=2$

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