To use trigonometric identities to find the value of $\tan\alpha$ I'm stuck on this problem:  
If $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha $ then find the value of $\tan\alpha$
I have tried isolating $\alpha$ but ended up getting $\tan\alpha = \frac{1}{\tan\beta}$ and $\tan\beta=\frac{1 - \tan\frac{\gamma}{2}}{1 + \tan\frac{\gamma}{2}}$   
Could someone please tell me how to properly approach this question?
 A: Given: 
$$\alpha+\beta=\frac{\pi}{2} \Rightarrow \alpha=\frac{\pi}{2}-\beta$$
$$\tan\alpha=\tan(\frac{\pi}{2}-\beta)=\cot \beta$$
$$\qquad=\frac{1}{\tan\beta}$$ giving us:
$$\tan\alpha\tan\beta=1$$
Because
$$ \tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$$
$$\tan\gamma=\frac{\tan\alpha-\tan\beta}{2}$$ $$\implies 2\tan\gamma=\tan\alpha-\tan\beta$$
$$\boxed{\tan\alpha=2\tan\gamma+\tan\beta} $$
A: $$\alpha+\beta=\dfrac\pi2$$
$$\alpha-\beta=\gamma$$
Add to find $\alpha$ in terms of $\gamma$
Then apply $$\tan(x+y)$$ formula
A: Note that in terms of $\tan \beta$
$$\alpha+\beta=\frac{\pi}{2} \Rightarrow \alpha=\frac{\pi}{2}-\beta\implies\tan \alpha=\frac{1}{\tan \beta}$$
and in terms of $\tan \gamma$
$$\beta+\gamma=\alpha\implies\beta=\alpha-\gamma$$
$$\alpha=\frac{\pi}{2}-\alpha+\gamma\implies2\alpha=\frac{\pi}{2}+\gamma\implies\tan 2\alpha=\frac{2\tan \alpha}{1-\tan^2 \alpha}=-\frac{1}{\tan \gamma}\quad \tan\gamma\neq0\\\implies\tan \alpha=\tan\gamma\pm{\sqrt{\tan^2\gamma+1}}$$
