Question

If $\cos \theta =\cos \alpha \cos \beta$, then $\tan \frac{\theta +\alpha }{2}\tan \frac{\theta -\alpha }{2}={\tan }^g\frac{\beta }{2}$
Find g

Answer 1

Given, $\cos \theta =\cos \alpha \cos \beta$. We have
$\cos \beta =\frac{\cos \theta }{\cos \alpha }$
Now, ${\tan }^2\frac{\beta }{2}=\frac{1-\cos \beta }{1+\cos \beta }=\frac{1-\frac{\cos \theta }{\cos \alpha }}{1+\frac{\cos \theta }{\cos \alpha }}$
$=\frac{\cos \alpha -\cos \theta }{\cos \alpha +\cos \theta }$
$=\frac{2\sin \frac{\alpha +\theta }{2}\sin \frac{\theta -\alpha }{2}}{2\cos \frac{\theta +\alpha }{2}\cos \frac{\theta -\alpha }{2}}=\tan \frac{\theta +\alpha }{2}\tan \frac{\theta -\alpha }{2}$
Ans: 2