Q60 / 160 · AP EAMCET 2016 · Mathematics

If $(1+\tan \alpha)(1+\tan 4 \alpha)=2, \alpha \in\left(0, \frac{\pi}{16}\right)$, then $\alpha$ is equal to

  • Correct answer
    A
    $\frac{\pi}{20}$
  • B
    $\frac{\pi}{30}$
  • C
    $\frac{\pi}{40}$
  • D
    $\frac{\pi}{60}$

Solution

$\begin{aligned} & \text { Given that }(1+\tan \alpha)(1+\tan 4 \alpha)=2, \alpha \in\left(0, \frac{\pi}{16}\right) \\ & \Rightarrow 1+\tan \alpha+\tan 4 \alpha+\tan \alpha \tan 4 \alpha=2 \\ & \Rightarrow \quad \tan \alpha+\tan 4 \alpha=1-\tan \alpha \tan 4 \alpha \\ & \Rightarrow \quad \frac{\tan \alpha+\tan 4 \alpha}{1-\tan \alpha+\tan 4 \alpha}=1 \\ & \Rightarrow \quad \tan (\alpha+4 \alpha)=1 \\ & \Rightarrow \quad \tan 5 \alpha=\tan \frac{\pi}{4} \\ & \Rightarrow \quad 5 \alpha=\frac{\pi}{4} \Rightarrow \tan \alpha=\frac{\pi}{20}\end{aligned}$

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

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