JEE Main 2021 (Online) 26th August Morning ShiftQuestion 1
The sum of solutions of the equation $\frac{\cos x}{1+\sin x}=|\tan2x|$, $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)-\left\{\frac{\pi}{4},-\frac{\pi}{4}\right\}$ is :
Trigonometric Ratio & Identities
Solution
Concept. $\frac{\cos x}{1+\sin x}=\tan\left(\frac\pi4-\frac x2\right)$; split by the sign of $\tan2x$. Formulas used. (1) $\dfrac{\cos x}{1+\sin x}=\tan\left(\dfrac\pi4-\dfrac x2\right)$(2) $\tan u=\tan v\iff u=v+k\pi$Step 1 - $\tan2x\ge0$. By (2), $2x=\frac\pi4-\frac x2+k\pi$, so $x=\frac{\pi}{10}+\frac{2k\pi}{5}$. In the interval: $x=\frac\pi{10}$ and $x=-\frac{3\pi}{10}$ (for both, $\tan2x>0$). Step 2 - $\tan2x<0$. $\tan\left(\frac\pi4-\frac x2\right)=\tan(-2x)$, so $x=-\frac\pi6-\frac{2k\pi}{3}$. In the interval: $x=-\frac\pi6$ (and $\tan\left(-\frac\pi3\right)<0$). Step 3 - Sum. $$\frac{\pi}{10}-\frac{3\pi}{10}-\frac{\pi}{6}=\frac{3\pi-9\pi-5\pi}{30}=-\frac{11\pi}{30}.$$ Answer: (B)