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

The Correct Option is C

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

Answer 2

Given :
\(S=\left\{\theta\in[0,2\pi]:8^{2\sin^2\theta}+8^{2\cos^2\theta}=16\right\}\)
Now apply AM ≥ GM for \(8^{2\sin^2\theta},8^{2\cos^2\theta}\)
\(\frac{8^{2\sin^2\theta}+8^{2\cos^2\theta}}{2}\ge\left(8^{2\sin^2\theta+2\cos^2\theta}\right)^{\frac{1}{2}}\)
8 ≥ 8
\(⇒8^{2\sin^2\theta}=8^{2\cos^2\theta}\) or \(\sin^2\theta=\cos^2\theta\)
\(\therefore\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\)
\(n(S)+\sum\limits_{\theta\in S}\sec(\frac{\pi}{4}+2\theta)\cosec(\frac{\pi}{4}+2\theta)\)
\(4+\sum\limits_{\theta\in S}\frac{2}{2\sin(\frac{\pi}{4}+2\theta)\cos(\frac{\pi}{4}+2\theta)}\)
\(=4+\sum\limits_{\theta\in S}\frac{2}{\sin(\frac{\pi}{2}+4\theta)}=4+2\sum\limits_{\theta\in S}\cosec(\frac{\pi}{2}+4\theta)\)
\(=4+2[\cosec(\frac{\pi}{2}+\pi)\cosec(\frac{\pi}{2}+3\pi)+\cosec(\frac{\pi}{2}+5\pi)+\cosec(\frac{\pi}{2}+7\pi)]\)
\(=4+2[-\cosec\frac{\pi}{2}-\cosec\frac{\pi}{2}-\cosec\frac{\pi}{2}-\cosec\frac{\pi}{2}]\)
\(=4-2(4)\)
\(=4-8=-4\)
So, the correct option is (C) : -4.