Question

Let \(S=\left\{0∈(0,\frac{π}{2}) : \sum^{9}_{m=1} \sec(θ+(m-1)\frac{π}{6})\sec(θ+\frac{mπ}{6}) = -\frac{8}{\sqrt3}\right\}\)
Then,

• A. \(S = \left\{\frac{π}{12}\right\}\)
• B. \(S = \left\{\frac{2π}{3}\right\}\)
• C. \(∑_{θ∈S}θ = \frac{π}{2}\)
• D. \(∑_{θ∈S}θ = \frac{3π}{4}\)

Answer 1

The Correct Option is C

The correct answer is (C) : \(∑_{θ∈S}θ = \frac{π}{2}\)
\(S=\left\{0∈(0,\frac{π}{2}) : \sum^{9}_{m=1} \sec(θ+(m-1)\frac{π}{6})\sec(θ+\frac{mπ}{6}) = -\frac{8}{\sqrt3}\right\}\)
\(∑^{9}_{m=1} \frac{1}{\cos(θ+(m-1)\frac{π}{6})}\cos(θ+m\frac{π}{6})\)
\(\frac{1}{\sin\left(\frac{\pi}{6}\right)} \sum_{m=1}^{9} \frac{\sin\left[\left(\theta + m\frac{\pi}{6}\right) - \left(\theta + (m-1)\frac{\pi}{6}\right)\right]}{\cos\left(\theta + (m-1)\frac{\pi}{6}\right)\cos\left(\theta + m\frac{\pi}{6}\right)}\)
\(2 \sum_{m=1}^{9} \left[ \tan\left(\theta + m\frac{\pi}{6}\right) - \tan\left(\theta + (m-1)\frac{\pi}{6}\right) \right]\)
Now, \(m=1,\ 2 \left[ \tan\left(\theta + \frac{\pi}{6}\right) - \tan(\theta) \right]\)
\(m=2\ \ \ \ \ 2 \left[ \tan\left(\theta + \frac{2\pi}{6}\right) - \tan\left(\theta + \frac{\pi}{6}\right) \right]\)
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.
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\(m = 9\ \ \ \ 2[\tan(θ+\frac{9π}{6})-\tan(θ+8\frac{π}{6})]\)
\(∴ = 2[\tan(θ+\frac{3π}{2})-\tanθ] = \frac{-8}{\sqrt3}\)
\(= -2[\cotθ+\tanθ] = \frac{-8}{\sqrt3}\)
\(= -\frac{2×2}{2\sinθ\cosθ} = \frac{-8}{\sqrt3}\)
\(= \frac{1}{2\sinθ} = \frac{2}{\sqrt3}\)
\(⇒ \sin2θ = \frac{\sqrt3}{2}\)
\(2θ = \frac{π}{3}\),
\(2θ = \frac{2π}{3}\)
\(θ = \frac{π}{6}\)
\(θ = \frac{π}{3}\)
\(∑θi = \frac{π}{6}+\frac{π}{3}\)
\(= \frac{π}{2}\)