Question

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

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

Answer 1

Answer: (A)

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