You already have a relation between $\theta$ and $\alpha$. If this is correct, all you need to do is to rearrange it. If it is not possible to rearrange it to get the given relation, then either your equation is wrong, or the given equation is wrong, or both.
If the particle grazes the upper plane at maximum height, where the velocity vector is horizontal, then the planes must be horizontal. The falling particle must then strike the lower plane vertically. This can only happen if the particle is projected vertically.
So grazing at maximum height requires $\theta = 0$ and $\alpha = 90^\circ$, no other combinations being possible. It seems your equation might be wrong.
However, perhaps you are using co-ordinate axes which are parallel and perpendicular to the inclined planes. In this frame of reference the particle does graze the upper plane at maximum height. The difficulty of this approach is transforming the equation of the trajectory to these axes - although, this may not be necessary.
Because time (eg time of flight, speed and $g$) do not feature in the answer, the problem can be solved using only co-ordinate geometry. Inverting the parabola and placing the vertex at the origin, the eqn of the trajectory is $4ay=x^2$ which has the parametric form $x=2at, y=at^2$ for some parameters $a>0$ and $-\infty < t < +\infty$.
The parabola is intersected by 3 parallel, equidistant lines at points with parameters $p, q, r$ where $q>p>0$ and $r<0$. The tangent to the parabola at a point with parameter p has slope p. The tangent at r is normal to that at p so $pr=-1$.
Using the fact that the lines are equidistant you should (I think) be able to express the tangent at $q=\tan(\theta + \alpha)$ in terms of $p=\tan\theta$, hence derive a relation between $\tan\theta$ and $\tan\alpha$.
