Q195 / 270 · NDA 2017 (Phase 1) · Mathematics

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

  • A
    -1
  • Correct answer
    B
    0
  • C
    1
  • D
    2

Solution

Given, $\sin \theta=3 \sin (\theta+2 \alpha)$ $\Rightarrow \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{1}{3}$ Apply componendo and dividendo rule $\begin{aligned} & \Rightarrow \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{1+3}{1-3} \\ & \Rightarrow \frac{2 \sin (\theta+\alpha) \cos \alpha}{2 \cos (\theta+\alpha) \sin \alpha}=\frac{4}{-2}=-2 \\ & \Rightarrow \frac{\tan (\theta+\alpha)}{\tan \alpha}=-2 \\ & \Rightarrow \tan (\theta+\alpha)=-2 \tan \alpha \Rightarrow \tan (\theta+\alpha)+2 \tan \alpha=0 \end{aligned}$

Hence, the correct answer is option: 'B'.

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