JEE Main 2022 (Online) 26th July Morning ShiftQuestion 1
Let $S=\{\theta\in[0,2\pi]:8^{2\sin^2\theta}+8^{2\cos^2\theta}=16\}$. Then $n(S)+\sum_{\theta\in S}\left(\sec\left(\frac\pi4+2\theta\right)\mathrm{cosec}\left(\frac\pi4+2\theta\right)\right)$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. With $t=8^{2\sin^2\theta}$ the condition is $(t-8)^2=0$; then evaluate each term. Formulas used. (1) $8^{2\cos^2\theta}=8^{2-2\sin^2\theta}=\frac{64}{t}$(2) $\sec x\operatorname{cosec}x=\dfrac{2}{\sin2x}$Step 1 - Solve. By (1), $t+\frac{64}{t}=16$, so $(t-8)^2=0$ and $2\sin^2\theta=1$. So $\theta=\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4$ and $n(S)=4$. Step 2 - Each term. By (2) with $x=\frac\pi4+2\theta$: $\sin2x=\sin\left(\frac\pi2+4\theta\right)=\cos4\theta$. For these $\theta$, $4\theta=\pi,3\pi,5\pi,7\pi$, so $\cos4\theta=-1$ and each term is $-2$. Step 3 - Value. $4+4(-2)=-4$. Answer: (C)