I am a graduate student, I was solving some high school problems for a student of mine. I am stuck on this one for quite a long time. (Any high school problem solving technique is usable)
$\tan^2 (Z) \cos^2 (A/2+B/2) = \sin A \sin B$
How to prove that,
$$\tan^2 (Z/2) = \tan (A/2) \tan (B/2)$$
Define Weierstrass variables $a=\tan\frac{A}{2}$ etc. so $\tan\frac{A+B}{2}=\frac{a+b}{1-ab}$ and $\sec^2\frac{A+B}{2}=\frac{(1+a^2)(1+b^2)}{(1-ab)^2}$. The given condition simplifies to $\frac{z^2}{(1-z^2)^2}=\frac{ab}{(1-ab)^2}$, whereas the desired result is $z^2=ab$. It therefore suffices to show $\frac{x}{(1-x)^2}$ is injective on $[0,\,1]$. Here's the graph.