Let $S=\{\theta \in[0,2 \pi]: 8^{2 \sin ^{2} \theta}+8^{2 \cos ^{2} \theta}=16\}$. 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 :

Trigonometric Equations — JEE Main · 2022 · 26 Jul 2022 · Shift 1 Question with Solution

Let

S = {θ \in [0, 2 π] : 8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 16}

. Then

n (S) + \sum_{θ \in S} (sec (\frac{π}{4} + 2 θ) cosec (\frac{π}{4} + 2 θ))

is equal to :

Answer: (C)

Rewrite the equation:

8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 8^{2 s i n^{2} θ} + 8^{2 (1 - s i n^{2} θ)} = 16

. Let

y = 8^{2 s i n^{2} θ}

, then

y + \frac{64}{y} = 16 \Rightarrow y^{2} - 16 y + 64 = 0 \Rightarrow (y - 8)^{2} = 0 \Rightarrow y = 8

. Thus

8^{2 s i n^{2} θ} = 8 \Rightarrow 2 sin^{2} θ = 1 \Rightarrow sin^{2} θ = \frac{1}{2} \Rightarrow sin θ = \pm \frac{1}{2}

. In

[0, 2 π]

θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}

, so

n (S) = 4

. For each

θ

sec (\frac{π}{4} + 2 θ) csc (\frac{π}{4} + 2 θ) = \frac{2}{s i n ( \frac{π}{2} + 4 θ )} = \frac{2}{c o s ( 4 θ )}

. For the four values,

4 θ = π, 3 π, 5 π, 7 π

, giving

cos (4 θ) = - 1

for all. Thus the sum is

4 \cdot \frac{2}{- 1} = - 8

, and

n (S) + sum = 4 - 8 = - 4

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option