Question

The sum of solutions of the equation $\frac{\cos x}{1+\sin x}=|\tan 2x|,x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)-\left\{\frac{\pi }{4},-\frac{\pi }{4}\right\}$ is :

• A. $-\frac{11\pi }{30}$
• B. $\frac{\pi }{10}$
• C. $-\frac{7\pi }{30}$
• D. $-\frac{\pi }{15}$

Answer 1

Answer: (A)

$\frac{\cos x}{1+\sin x}=|\tan 2x|$
$\Rightarrow \frac{{\cos }^{2}x/2-{\sin }^{2}x/2}{(\cos x/2+\sin x/2)}=|\tan 2x|$
$\Rightarrow {\tan }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)={\tan }^{2}2x$
$\Rightarrow 2x=n\pi \pm \left(\frac{\pi }{4}-\frac{x}{2}\right)$
$\Rightarrow x=\frac{-3\pi }{10},\frac{-\pi }{6},\frac{\pi }{10}$
or sum $=\frac{-11\pi }{6}$