If $\alpha +\beta =\frac{\pi }{2}$ and $\beta +\gamma =\alpha$, then the value of $\tan \alpha$ is

The correct option is C

$\tan \beta +2\tan \gamma$

Explanation for the correct option:

Step 1. Find the value of $\tan \alpha$:

Given, $\alpha +\beta =\frac{\pi }{2}$ and $\beta +\gamma =\alpha$

$\Rightarrow$ $\gamma =\alpha –\beta$

Step 2. Take “tan” on both sides, we get

$\begin{array}{rcl}\tan \gamma & =& \tan (\alpha –\beta )\\ & =& \frac{\tan \alpha –\tan \beta }{1+\tan \alpha \tan \beta }\\ & =& \frac{\tan \alpha –\tan \beta }{1+\tan \alpha \tan \left({\displaystyle \frac{\pi }{2}}–\alpha \right)}\\ & =& \frac{\tan \alpha –\tan \beta }{1+\tan \alpha cot\alpha }\\ & =& \frac{\tan \alpha –\tan \beta }{1+1}\end{array}$

$\Rightarrow 2\tan \gamma =\tan \alpha –\tan \beta$

$\mathbf{\therefore }\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\alpha }\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\beta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\gamma }$

Hence, Option ‘C’ is Correct.