JEE Main 2022 · 26 Jul · Shift 1 · Mathematics Q18

← JEE Main 2022 · 26 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 18 of 90

Let $S=\left\{\theta\in[0,2\pi]:8^{2\sin^2\theta}+8^{2\cos^2\theta}=16\right\}$ . Then $n(S)+\displaystyle\sum_{\theta\in S}\left(\sec\left(\frac\pi4+2\theta\right)\operatorname{cosec}\left(\frac\pi4+2\theta\right)\right)$ is equal to :

-2

-4

Answer and worked solution

Official answer: C

Equation. With $t=8^{2\sin^2\theta}$ : $t+\frac{64}t=16$ , $t=8$ , so $\sin^2\theta=\frac12$ : four values of θ.

Each term. $\sec u\,\operatorname{cosec}u=\frac2{\sin2u}$ with $2u=\frac\pi2+4\theta$ : $\frac2{\cos4\theta}=\frac2{-1}=-2$ . Sum $-8$ .

Total $4-8=-4$ .

(C)

Topics:Trigonometric equations

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JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 18