MCQ
If $\alpha+\beta=\frac{\pi}{2}$and$\beta+\gamma=\alpha$, then $\tan\alpha$ equals
  • A
    $2(\tan\beta+\tan\gamma)$
  • B
    $\tan\beta+\tan\gamma$
  • $\tan\beta+2\tan\gamma$
  • D
    $2\tan\beta+\tan\gamma$

Answer

Correct option: C.
$\tan\beta+2\tan\gamma$
(C)
$\beta+\gamma=\alpha$
$\Rightarrow\gamma=\alpha-\beta$
$\Rightarrow\tan\gamma=\tan(\alpha-\beta)$
$\Rightarrow\tan\gamma=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$
$\Rightarrow\tan\gamma=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\cot\alpha}$
$\ldots\left[\because\alpha+\beta=\frac{\pi}{2},\therefore\beta=\frac{\pi}{2}-\alpha\right]$
$\Rightarrow\tan\gamma=\frac{1}{2}(\tan\alpha-\tan\beta)$
$\Rightarrow\tan\alpha=\tan\beta+2\tan\gamma$

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