Explanation

(c)
Given, $\alpha + \beta = \frac{\pi}{2},\ \ \beta + \gamma = \alpha$
$\Rightarrow \ \ \beta = \frac{\pi}{2} - \alpha,\ \ \beta + \gamma = \alpha$
$\Rightarrow \ \tan{\beta = \tan\left( \frac{\pi}{2} - \alpha \right)}$ and $\tan{(\beta + \gamma) = \tan\alpha}$
$\Rightarrow \ \tan{\beta = \cot\alpha}$ …(i)
And $\frac{\tan{\beta + \tan\gamma}}{1 - \tan{\beta\tan\gamma}} = \tan\alpha$
$\Rightarrow \ \ \ \frac{\tan{\beta + \tan\gamma}}{1 - {cot\ \alpha}{\tan\gamma}} = \frac{\tan\alpha}{1}$ [from Eq. (i)]
$\Rightarrow \ \tan{\beta + \tan{\gamma = \tan{\alpha - \tan\gamma}}}$
$\Rightarrow \ \tan{\alpha = \tan{\beta + 2\tan\gamma}}$