Question

If $\cos \theta =\frac{a\cos \varphi +b}{a+b\cos \varphi }$, prove that $\tan \left(\theta /2\right)=\sqrt{\left[(a-b)/(a+b)\right]}\tan \left(\varphi /2\right)$

Answer 1

$\cos \theta =\frac{a\cos \varphi }{a+b\cos \varphi }$

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

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

Apply componendo and dividendo

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

$\tan \left(\theta /2\right)=\sqrt{(a-b)/(a+b)}\tan \left(\varphi /2\right)$