Q198 / 270 · NDA 2017 (Phase 1) · Mathematics

If $\tan (\alpha+\beta)=2$ and $\tan (\alpha-\beta)=1$, then $\tan (2 \alpha)$ is equal to

  • Correct answer
    A
    -3
  • B
    -2
  • C
    -1/3
  • D
    1

Solution

$\begin{aligned} & \tan (\alpha+\beta)=2 \\ & \tan (\alpha-\beta)=1 \\ & \tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)] \\ & =\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)} \\ & =\frac{2+1}{1-2.1}=\frac{3}{-1}=-3\end{aligned}$

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

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