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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2017 · Mathematics · Trigonometric Ratios & Identities

If θ=3 (θ+2 α) , then the value of (θ+α)+2 α is equal to

If sin⁡θ=3sin⁡(θ+2α)\sin \theta=3 \sin (\theta+2 \alpha), then the value of tan⁡(θ+α)+2tan⁡α\tan (\theta+\alpha)+2 \tan \alpha is equal to

Options

  1. A-1
  2. B0
  3. C1
  4. D2

Correct answer

B. 0

Step-by-step solution

Given, sin⁡θ=3sin⁡(θ+2α)⇒sin⁡(θ+2α)sin⁡θ=13\sin \theta=3 \sin (\theta+2 \alpha)\Rightarrow \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{1}{3} Apply componendo and dividendo rule ⇒sin⁡(θ+2α)+sin⁡θsin⁡(θ+2α)−sin⁡θ=1+31−3⇒2sin⁡(θ+α)cos⁡α2cos⁡(θ+α)sin⁡α=4−2=−2⇒tan⁡(θ+α)tan⁡α=−2⇒tan⁡(θ+α)=−2tan⁡α⇒tan⁡(θ+α)+2tan⁡α=0\begin{aligned} & \Rightarrow \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{1+3}{1-3} \\ & \Rightarrow \frac{2 \sin (\theta+\alpha) \cos \alpha}{2 \cos (\theta+\alpha) \sin \alpha}=\frac{4}{-2}=-2 \\ & \Rightarrow \frac{\tan (\theta+\alpha)}{\tan \alpha}=-2 \\ & \Rightarrow \tan (\theta+\alpha)=-2 \tan \alpha \Rightarrow \tan (\theta+\alpha)+2 \tan \alpha=0 \end{aligned}
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