Solveeit Logo
Login

Question

Mathematics Question on Trigonometric Equations

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∈Sθ=π2∑_{θ∈S}θ = \frac{π}{2}

D

∑θ∈Sθ=3π4∑_{θ∈S}θ = \frac{3π}{4}

Answer

∑θ∈Sθ=π2∑_{θ∈S}θ = \frac{π}{2}

Explanation

Solution

The correct answer is (C) : ∑θ∈Sθ=π2∑_{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\\}
∑m=191cos⁡(θ+(m−1)π6)cos⁡(θ+mπ6)∑^{9}_{m=1} \frac{1}{\cos(θ+(m-1)\frac{π}{6})}\cos(θ+m\frac{π}{6})
1sin⁡(π6)∑m=19sin⁡[(θ+mπ6)−(θ+(m−1)π6)]cos⁡(θ+(m−1)π6)cos⁡(θ+mπ6)\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∑m=19[tan⁡(θ+mπ6)−tan⁡(θ+(m−1)π6)]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[tan⁡(θ+π6)−tan⁡(θ)]m=1,\ 2 \left[ \tan\left(\theta + \frac{\pi}{6}\right) - \tan(\theta) \right]
m=2     2[tan⁡(θ+2π6)−tan⁡(θ+π6)]m=2\ \ \ \ \ 2 \left[ \tan\left(\theta + \frac{2\pi}{6}\right) - \tan\left(\theta + \frac{\pi}{6}\right) \right]
.
.
.
m=9    2[tan⁡(θ+9π6)−tan⁡(θ+8π6)]m = 9\ \ \ \ 2[\tan(θ+\frac{9π}{6})-\tan(θ+8\frac{π}{6})]
∴=2[tan⁡(θ+3π2)−tan⁡θ]=−83∴ = 2[\tan(θ+\frac{3π}{2})-\tanθ] = \frac{-8}{\sqrt3}
=−2[cot⁡θ+tan⁡θ]=−83= -2[\cotθ+\tanθ] = \frac{-8}{\sqrt3}
=−2×22sin⁡θcos⁡θ=−83= -\frac{2×2}{2\sinθ\cosθ} = \frac{-8}{\sqrt3}
=12sin⁡θ=23= \frac{1}{2\sinθ} = \frac{2}{\sqrt3}
⇒sin⁡2θ=32⇒ \sin2θ = \frac{\sqrt3}{2}
2θ=π32θ = \frac{π}{3},
2θ=2π32θ = \frac{2π}{3}
θ=π6θ = \frac{π}{6}
θ=π3θ = \frac{π}{3}
∑θi=π6+π3∑θi = \frac{π}{6}+\frac{π}{3}
=π2= \frac{π}{2}