Question

Prove that $2{\tan }^{-1}(\sqrt{\frac{(a-b)}{a+b}}\tan \frac{\theta }{2})={\cos }^{-1}[\frac{a\cos \theta +b}{a+b\cos \theta }]$.

Answer 1

$2{\tan }^{-1}(\sqrt{\frac{(a-b)}{a+b}}\tan \frac{\theta }{2})$

$={\cos }^{-1}\left(\frac{1-\frac{(a-b){\tan }^2\frac{\theta }{2}}{(a+b)}}{1+\frac{(a-b){\tan }^2\frac{\theta }{2}}{(a+b)}}\right)$ [Using formula]

$={\cos }^{-1}\left(\frac{(a+b)-(a-b){\tan }^2\frac{\theta }{2}}{(a+b)+(a-b){\tan }^2\frac{\theta }{2}}\right)$

$={\cos }^{-1}\left(\frac{(a+b){\cos }^2\frac{\theta }{2}-(a-b){\sin }^2\frac{\theta }{2}}{(a+b){\cos }^2\frac{\theta }{2}+(a-b){\sin }^2\frac{\theta }{2}}\right)$

$={\cos }^{-1}\left[\frac{a\cos \theta +b}{a+b\cos \theta }\right]$.