Let $S=\left\{\theta \in\left(0, \frac{\pi}{2}\right): \sum_{m=1}^{9} \sec \left(\theta+(m-1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{m \pi}{6}\right)=-\frac{8}{\sqrt{3}}\right\}$ Then

(C) $\sum_{\theta \in S} \theta=\frac{\pi}{2}$

(A) $S=\left\{\frac{\pi}{12}\right\}$

(B) $S=\left\{\frac{2 \pi}{3}\right\}$

(D) $\sum_{\theta \in S} \theta=\frac{3 \pi}{4}$

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Trigonometric Ratios and Identities

Let

S = {θ \in (0, \frac{π}{2}) : \sum_{m = 1}^{9} sec (θ + (m - 1) \frac{π}{6}) sec (θ + \frac{mπ}{6}) = - \frac{8}{3}}

Then

Answer: (C)

Using the telescoping sum technique with

β - α = π /6

, the sum simplifies to

2 (tan (θ + 3 π /2) - tan θ) = 2 (- cot θ - tan θ) = - \frac{8}{3}

. This gives

tan θ + cot θ = \frac{4}{3}

. Solving:

tan^{2} θ - \frac{4}{3} tan θ + 1 = 0

, which yields

tan θ = \frac{1}{3}

3

. Thus

θ = \frac{π}{6}

\frac{π}{3}

, so

S = {\frac{π}{6}, \frac{π}{3}}

and

\sum_{θ \in S} θ = \frac{π}{2}

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option