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Trigonometrical Equations
hard
Let,$S=\left\{\theta \in[0,2 \pi]: 8^{2 \sin ^{2} \theta}+8^{2 \cos ^{2} \theta}=16\right\}$. Then $n ( S )+\sum_{\theta \in S}\left(\sec \left(\frac{\pi}{4}+2 \theta\right) \operatorname{cosec}\left(\frac{\pi}{4}+2 \theta\right)\right)$ is equal to.
A
$0$
B
$-2$
C
$-4$
D
$12$
(JEE MAIN-2022)
Solution
$8^{2 \sin ^{2} \theta}+8^{2-2 \sin ^{2} \theta}=16$
$y+\frac{64}{y}=16$
$\Rightarrow y =8$
$\Rightarrow \sin ^{2} \theta=1 / 2$
$n ( S )+\sum_{\theta \in S} \frac{1}{\cos (\pi / 4+2 \theta) \sin (\pi / 4+2 \theta)}$
$=4+(-2) \times 4=-4$
Standard 11
Mathematics
