Question

If $\tan \frac{x-y}{2},\tan z,\tan \frac{x+y}{2}$ are in G.P., then show that $\cos x=\cos y.\cos 2z$

Answer 1

$b^2=ac\Rightarrow \frac{{\sin }^{2}z}{{\cos }^{2}z}=\frac{\sin \frac{x+y}{2}\sin \frac{x-y}{2}}{\cos \frac{x+y}{2}\cos \frac{x-y}{2}}$
$\frac{{\sin }^{2}z}{{\cos }^{2}z}=\frac{\cos y-\cos x}{\cos y+\cos x}$
cross multiply or by $C$ and $D$
$\cos x.1=\cos y({\cos }^{2}z-{\sin }^{2}z)=\cos y\cos 2z$

Hence proved.