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.

Alternatively:
$$\cos a=2\cos^2 \frac a2-1; \\
\cos^2 \frac a2=\frac 1{1+\tan^2 \frac a2}=\frac1{1+s^2}\\
\cos a+\cos b=2\cos ^2 \frac a2+2\cos^2 \frac b2-2=\\
\frac2{1+s^2}+\frac2{1+t^2}-2=1.$$
I leave to you to simplify it.
