Question

If $m{\tan }^{-1}\left(\sqrt{\frac{(a-b)}{a+b}}\tan \frac{\theta }{2}\right)={\cos }^{-1}\left[\frac{a\cos \theta +b}{a+b\cos \theta }\right]$.Find $m$

Answer 1

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

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

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

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

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

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

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