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Mathematics Question on Trigonometry

Let S=\left\\{θ∈[0,2π]:8^{2sin^2⁡θ}+8^{2cos^2⁡θ}=16\right\\} .
Thenn(S)+∑θ∈S(sec⁡(π4+2θ)cosec⁡(π4+2θ))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

-4

Explanation

Solution

S=\left\\{θ∈[0,2π]:8^{2\sin^2⁡θ}+8^{2\cos^2⁡θ}=16\right\\}
Now apply AM≥GM for82sin⁡2⁡θ,82cos⁡2⁡θAM≥ GM\ for 8^{2\sin^2⁡θ},8^{2\cos^2⁡θ}
82sin⁡2⁡θ+82cos⁡2⁡θ2≥(82sin⁡2⁡θ+2cos⁡2⁡θ)12\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≥88≥8
⇒82sin⁡2⁡θ=82cos⁡2⁡θ⇒8^{2\sin^2⁡θ}=8^{2\cos^2⁡θ}
sin⁡2θ=cos⁡2θ\sin^2θ=\cos^2θ
∴θ=π4,3π4,5π4,7π4∴θ=\frac{π}{4},\frac{3π}{4},\frac{5π}{4},\frac{7π}{4}
n(S)+∑θ∈Ssec⁡⁡(π4+2θ)cosec⁡(π4+2θ)n(S)+\sum_{\theta∈S}\sec⁡(\frac{π}{4}+2θ)\cosec(\frac{π}{4}+2θ)
4+∑θ∈S22sin⁡⁡(π4+2θ)cos⁡⁡(π4+2θ)4+\sum_{θ∈S} \frac{2}{2\sin⁡(\frac{π}{4}+2θ)\cos⁡(\frac{π}{4}+2θ)}
=4+∑θ∈S2sin⁡⁡(π2+4θ)=4+2∑θ∈Scosec⁡(π2+4θ)=4+\sum_{θ∈S} \frac{2}{\sin⁡(\frac{π}{2}+4θ)}=4+2\sum_{θ∈S}\cosec(\frac{π}{2}+4θ)
=4+2[cosec⁡(π2+π)+cosec⁡(π2+3π)+.cosec⁡(π2+5π)+cosec⁡(π2+7π)]=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⁡π2−cosec⁡π2−cosec⁡π2−cosec⁡π2]=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