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The sum of solutions of the equation $\frac{\cos \mathrm{x}}{1+\sin \mathrm{x}}=|\tan 2 \mathrm{x}|, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)-\left\{\frac{\pi}{4},-\frac{\pi}{4}\right\}$ is :
A
$-\frac{11 \pi}{30}$
B
$\frac{\pi}{10}$
C
$-\frac{7 \pi}{30}$
D
$-\frac{\pi}{15}$
(JEE MAIN-2021)
Solution
$\frac{\cos x}{1+\sin x}=|\tan 2 x|$
$\Rightarrow \frac{\cos ^{2} x / 2-\sin ^{2} x / 2}{(\cos x / 2+\sin x / 2)}=|\tan 2 x|$
$\Rightarrow \tan ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan ^{2} 2 x$
$\Rightarrow 2 x=n \pi \pm\left(\frac{\pi}{4}-\frac{x}{2}\right)$
$\Rightarrow x=\frac{-3 \pi}{10}, \frac{-\pi}{6}, \frac{\pi}{10}$
$\text { or sum }=\frac{-11 \pi}{30}$
Standard 11
Mathematics
