Question

If $\cos \theta =\cos \alpha \cos \beta$, prove that $\tan \frac{\theta +\alpha }{2}\tan \frac{\theta -\alpha }{2}={\tan }^2\frac{\beta }{2}$

Answer 1

$\cos \theta =\cos \alpha \cos \beta$

$\therefore \cos \beta =\frac{\cos \theta }{\cos \alpha }$ ......$\left(1\right)$

R.H.S$={\tan }^2\frac{\beta }{2}=\frac{{(1-\cos \beta )}^2}{{\sin }^2\beta }$

$\Rightarrow \frac{{(1-\cos \beta )}^2}{1-{\cos }^2\beta }=\frac{{(1-\cos \beta )}^2}{(1-\cos \beta )(1+\cos \beta )}$

$\Rightarrow \frac{(1-\cos \beta )}{(1+\cos \beta )}=\frac{1-\frac{\cos \theta }{\cos \alpha }}{1+\frac{\cos \theta }{\cos \alpha }}$

$=\frac{\cos \alpha -\cos \theta }{\cos \alpha +\cos \theta }$

$=\frac{2\sin \left(\frac{\alpha +\theta }{2}\right)\sin \left(\frac{\alpha -\theta }{2}\right)}{2\cos \left(\frac{\alpha +\theta }{2}\right)\sin \left(\frac{\theta -\alpha }{2}\right)}$

$=\tan \left(\frac{\theta +\alpha }{2}\right)\tan \left(\frac{\theta -\alpha }{2}\right)$

$=L.H.S$

Hence proved.