Question

$\text{If}\tan \frac{\theta }{2}=\sqrt{\frac{a-b}{a+b}}\tan \frac{\varphi }{2}\text{prove that}\cos \theta =\frac{a\cos \varphi +b}{a+b\cos \varphi }$

Answer 1

We know that,$\cos \theta =\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\Rightarrow \cos \theta =\frac{1-\frac{a-b}{a+b}{\tan }^2\frac{\varphi }{2}}{1+\frac{a-b}{a+b}{\tan }^2\frac{\varphi }{2}}[\because \tan \frac{\theta }{2}=\sqrt{\frac{a-b}{a+b}}\tan \frac{\varphi }{2}]\cos \theta =\frac{a+b-a{\tan }^2\frac{\varphi }{2}+b{\tan }^2\frac{\varphi }{2}}{a+b+a{\tan }^2\frac{\varphi }{2}-b{\tan }^2\frac{\varphi }{2}}\Rightarrow \cos \theta =\frac{a(1-{\tan }^2\frac{\varphi }{2})+(1+{\tan }^2\frac{\varphi }{2})}{a(1+{\tan }^2\frac{\varphi }{2})+b(1-{\tan }^2\frac{\varphi }{2})}\text{On dividing RHS by 1 +}{\tan }^2\frac{\varphi }{2}\text{we get},\cos \theta =\frac{a\left(\frac{1-ta{n}^2\frac{\varphi }{2}}{1+{\tan }^2\frac{\varphi }{2}}\right)+b}{a+b\left(\frac{1-{\tan }^2\frac{\varphi }{2}}{1+{\tan }^2\frac{\varphi }{2}}\right)}=\frac{a\cos \varphi +b}{a+b\cos \theta }$