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

Given $\alpha+\beta=\frac{\pi}{2}$, we get $\alpha=\frac\pi2-\beta$.<br><br>Also, $\beta+\gamma=\alpha$. Substituting for $\alpha$ gives $$\beta+\gamma=\frac\pi2-\beta$$ so $$\gamma=\frac\pi2-2\beta.$$<br><br>Now $$\tan\alpha=\tan(\beta+\gamma).$$ Using the addition formula, $$\tan\alpha=\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}.$$<br><br>From $\alpha+\beta=\frac\pi2$, we have $$\tan\alpha\tan\beta=1.$$ Let $x=\tan\beta$ and $y=\tan\gamma$. Then $$\frac1x=\frac{x+y}{1-xy}.$$ This gives $$1-xy=x^2+xy$$ and hence $$1=x^2+2xy.$$ Dividing by $x$ gives $$\frac1x=x+2y.$$ Therefore $$\tan\alpha=\tan\beta+2\tan\gamma.$$<br><br>This matches option $\text{C}$.