<s>[INST] 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.
 [/INST] $\frac{1-t^2}{1+t^2} +\frac{1-s^2}{1+s^2}=1$
$1+s^2-t^2-s^2t^2+1+t^2-s^2-s^2t^2=1+s^2+t^2+s^2t^2$
$\Rightarrow 1-s^2-t^2-3s^2t^2=0$
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