Question

If $\tan \beta =\cos \theta \tan \alpha$, then prove that $\sin (\alpha -\beta )={\tan }^2\left(\theta /2\right)\sin (\alpha +\beta )$

Answer 1

The given relation is $\tan \alpha /\tan \beta =1/\cos \theta$
Apply componendo and dividendo. Then
$\frac{\tan \alpha -\tan \beta }{\tan \alpha +\tan \beta }=\frac{1-\cos \theta }{1+\cos \theta }$
or $\frac{\sin (\alpha -\beta )}{\sin (\alpha +\beta )}=\frac{2{\sin }^2\left(\theta /2\right)}{2{\cos }^2\left(\theta /2\right)}={\tan }^2\frac{\theta }{2}$ etc

Hence proved