A shot is fired with a velocity  $u$ at an angle $(\alpha +\theta )$ with the horizontal from the foot of an incline plane of angle $\alpha$  through the point of projection. If it hits the plane horizontally then 

The vertical velocity at P is zero
Hence,  $v\sin (\theta +\alpha )-gt=0$
$\Rightarrow t=\frac{v\sin (\theta +\alpha )}{g}\mathrm{.....}\left(1\right)$

Since displacement perpendicular to the plane is zero hence
$o=\left(v\sin \theta \right)t+\frac{1}{2}(-g\cos \alpha )t^2$
$t=\frac{2v\sin \theta }{g\cos \alpha }\mathrm{.....}\left(2\right)$
Equating (1) &(2) and rearranging, we get
$\frac{{\displaystyle v\sin \left(\theta +\alpha \right)}}{{\displaystyle g}}=\frac{{\displaystyle 2v\sin \theta }}{{\displaystyle g\cos \alpha }}$
$$\begin{gathered} \Rightarrow 2\sin \theta =\cos \alpha (\sin \theta \cos \alpha +\cos \theta \sin \alpha ) \\ \Rightarrow 2\sin \theta -{\cos }^2\alpha \sin \theta +\cos \theta \sin \alpha \cos \alpha \\ =(2-{\cos }^2\alpha )\sin \theta =\cos \theta \sin \alpha \cos \alpha \\ \Rightarrow \tan \theta =\frac{\sin \alpha \cos \alpha }{2-{\cos }^2\alpha } \\ \Rightarrow \tan \theta =\frac{\sin \alpha \cos \alpha }{(1-{\cos }^2\alpha )+1}=\frac{\sin \alpha \cos \alpha }{1+{\sin }^2\alpha } \\ \Rightarrow \tan \theta =\frac{\tan \alpha }{1+2{\tan }^2\alpha } \end{gathered}$$