If the coordinates of vertex of the triangle are (x,y), then

$\tan \alpha =$ $\cfrac{y}{x}$; $\tan \beta =$ $\cfrac{y}{R-x}$, where $R$ is the range of the projectile ($\cfrac{u^2 \sin 2\theta}{g}$). $\implies$ $\tan \alpha + \tan \beta = $ $\cfrac{y}{x} \cfrac{R}{R-x}$

Now, $x = u \cos \theta t$; $y = u \sin \theta t – 1/2 g t^2$, where $t$ is the time when projectile is at (x,y)

Substituting the expression for y, x and R in the equation for $\tan \alpha + \tan \beta$, we will get $\tan \theta$ on the right hand side