Given $\cos(a) +\cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$ Given $\cos(a) + \cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s  = \tan(a/2)$ and $t  = \tan(b/2)$.
I have tried using the identity $\cos(a) = \frac{1-t^2}{1+t^2}$. but manipulating this seems to have got me nowhere.

Hint:
$$\dfrac{1-s^2}{1+s^2}+\dfrac{1-t^2}{1+t^2}=1$$
$$\iff\dfrac{1-s^2}{1+s^2}=1-\dfrac{1-t^2}{1+t^2}$$
$$\iff(1-s^2)(1+t^2)=2t^2(1+s^2)$$
