Question

Let S =\left\\{\theta \in[0,2 \pi]: 8^{2 \sin ^2 \theta}+8^{2 \cos ^2 \theta}=16\right\\} Then $n ( S )+\displaystyle\sum_{\theta \in S }\left(\sec \left(\frac{\pi}{4}+2 \theta\right) \operatorname{cosec}\left(\frac{\pi}{4}+2 \theta\right)\right)$ is equal to:

• A. 0
• B. $-2$
• C. -4
• D. 12

Answer 1

Answer: (C)

-4

Answer 2

$8^{2\sin^2\theta}+8^{2-2\sin^2\theta}=16$
$y+\frac{64}{y}=16$
$⇒y=8$
$⇒\sin^2\theta=\frac{1}{2}$
$\therefore n(S)+\sum_{\theta\ ∈\ S}\frac{1}{\cos(\frac{\pi}{4}+2\theta)\sin(\frac{\pi}{4}+2\theta)}$
$=4+(-2)\times4$
$=-4$
So, the correct option is (C) : -4.