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:

Trigonometric Equations — JEE Main · 2021 · 26 Aug 2021 · Shift 1 Question with Solution

The sum of solutions of the equation

\frac{c o s x}{1 + s i n x} = ∣ tan 2 x ∣, x \in (- \frac{π}{2}, \frac{π}{2}) - {\frac{π}{4}, - \frac{π}{4}}

is:

Answer: (D)

Half-angle reduction gives

\frac{cos x}{1 + sin x} = \frac{cos ^{2} \frac{x}{2} - sin ^{2} \frac{x}{2}}{( cos \frac{x}{2} + sin \frac{x}{2} ) ^{2}} = \frac{1 - tan \frac{x}{2}}{1 + tan \frac{x}{2}} = tan (\frac{π}{4} - \frac{x}{2}) .

For

x \in (- \frac{π}{2}, \frac{π}{2})

the angle

\frac{π}{4} - \frac{x}{2}

lies in

(0, \frac{π}{2})

, so the left side is positive and the modulus on the right must be resolved by cases.
Where

tan 2 x \geq 0

the equation reads

\frac{π}{4} - \frac{x}{2} = 2 x + nπ

, giving

x = \frac{π}{10} - \frac{2 nπ}{5}

; inside the domain this yields

x = \frac{π}{10}

and

x = - \frac{3 π}{10}

, and at both of these

tan 2 x > 0

.
Where

tan 2 x < 0

the equation reads

\frac{π}{4} - \frac{x}{2} = - 2 x + nπ

, giving

x = \frac{2 nπ}{3} - \frac{π}{6}

; inside the domain this yields only

x = - \frac{π}{6}

, where

tan 2 x = - 3

.
Adding the three solutions,

\frac{π}{10} - \frac{3 π}{10} - \frac{π}{6} = - \frac{π}{5} - \frac{π}{6} = - \frac{11 π}{30} .

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option