A particle is projected at speed $u$, at angle $\alpha$, from level ground. $\theta$ is the angle of elevation of the projectile from the initial point of projection and $\phi$ is the angle that the velocity of the projectile makes with the horizontal. Show that $\tan\alpha+\tan\phi=2\tan\theta$ at all points on the trajectory.
My attempt:
The equations for the motion of the projectile are $$\ddot{x}=0, \ddot{y}=-g, \dot{x}=u\cos\alpha, \dot{y}=u\sin\alpha-gt, x=tu\cos\alpha, y=tu\sin\alpha-\dfrac{1}{2}gt^2$$
$$\begin{aligned}\tan\theta&=\dfrac{y}{x}=\tan\alpha-\dfrac{g}{2u}\sec\alpha \quad (\star)\\ \tan\phi&=\dfrac{\dot{y}}{\dot{x}}=\tan\alpha-\frac{gt}{2u}\sec\alpha\\ \end{aligned}$$
Rearranging the second equation and substituting into $(\star)$, we get $$\tan\theta=\tan\alpha-\dfrac{\tan\alpha-\tan\phi}{t}$$ or $$t\tan\theta=(t-1)\tan\alpha+\tan\phi$$ Now substituting $t=2$ gives $$2\tan\theta=\tan\alpha+\tan\phi.$$ But this does not seem quite right - I have only showed it's true when $t=2$. Is my approach wrong?