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# If $\alpha+\beta=\large\frac{\pi}{2}$ and $\beta+\gamma=\alpha$ then $\tan \alpha$ equals

$\begin{array}{1 1}(a)\;2(\tan\beta+\tan\gamma)&(b)\;\tan\beta+\tan\gamma\\(c)\;\tan\beta+2\tan\gamma&(d)\;2\tan\beta+\tan\gamma\end{array}$

Can you answer this question?

Given
$\alpha+\beta=\large\frac{\pi}{2}$
$\Rightarrow \alpha=\large\frac{\pi}{2}$$-\beta \tan\alpha=\tan\big(\large\frac{\pi}{2}$$-\beta\big)$
$\qquad=\cot \beta$
$\qquad=\large\frac{1}{\tan\beta}$
$\tan\alpha\tan\beta=1$
$\Rightarrow 1+\tan\alpha\tan\beta=2$
$\therefore \tan(\alpha-\beta)=\large\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$
$\tan\gamma=\large\frac{\tan\alpha-\tan\beta}{2}$
$2\tan\gamma=\tan\alpha-\tan\beta$
$\tan\alpha=2\tan\gamma+\tan\beta$
Hence (c) is the correct answer.
answered Oct 14, 2013