Question

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

• A. 0
• B. -2
• C. -4
• D. 12

Answer 1

The Correct Option is C

\(S=\left\{θ∈[0,2π]:8^{2\sin^2⁡θ}+8^{2\cos^2⁡θ}=16\right\}\)
Now apply \(AM≥ GM\ for 8^{2\sin^2⁡θ},8^{2\cos^2⁡θ}\)
\(\frac{8^{2\sin^2⁡θ}+8^{2\cos^2⁡θ}}{2}≥(8^{2\sin^2⁡θ}+2^{\cos^2⁡θ})^{\frac{1}{2}}\)
\(8≥8\)
\(⇒8^{2\sin^2⁡θ}=8^{2\cos^2⁡θ}\)
\(\sin^2θ=\cos^2θ\)
\(∴θ=\frac{π}{4},\frac{3π}{4},\frac{5π}{4},\frac{7π}{4}\)
\(n(S)+\sum_{\theta∈S}\sec⁡(\frac{π}{4}+2θ)\cosec(\frac{π}{4}+2θ)\)
\(4+\sum_{θ∈S} \frac{2}{2\sin⁡(\frac{π}{4}+2θ)\cos⁡(\frac{π}{4}+2θ)}\)
\(=4+\sum_{θ∈S} \frac{2}{\sin⁡(\frac{π}{2}+4θ)}=4+2\sum_{θ∈S}\cosec(\frac{π}{2}+4θ)\)
\(=4+2[\cosec(\frac{π}{2}+π)+\cosec(\frac{π}{2}+3π)+.\cosec(\frac{π}{2}+5π)+\cosec(\frac{π}{2}+7π)]\)
\(=4+2[−\cosec\frac{π}{2}−\cosec\frac{π}{2}−\cosec\frac{π}{2}−\cosec\frac{π}{2}]\)
= 4– 2(4)
= 4 – 8
= – 4
So, the correct option is (C): -4