Question

If $\cos \theta =\frac{a\cos \varphi +b}{a+b\cos \varphi }$, then $\tan \frac{\theta }{2}$ is equal to?

• A. $\sqrt{\left(\frac{a-b}{a+b}\right)}\tan \frac{\varphi }{2}$
• B. $\sqrt{\left(\frac{a+b}{a-b}\right)}\cos \frac{\varphi }{2}$
• C. $\sqrt{\left(\frac{a-b}{a+b}\right)}\sin \frac{\varphi }{2}$
• D. None of these

Answer 1

The correct option is A $\sqrt{\left(\frac{a-b}{a+b}\right)}\tan \frac{\varphi }{2}$

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

we know that $\cos A=\frac{1-{\tan }^2\frac{A}{2}}{1+{\tan }^2\frac{A}{2}}\Rightarrow \frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{a\cos \varphi +b}{a+b\cos \varphi }$

$\Rightarrow a+b\cos \varphi -a{\tan }^2\frac{\theta }{2}-b\cos \varphi {\tan }^2\frac{\theta }{2}=a\cos \varphi +b+a{\tan }^2\frac{\theta }{2}\cos \varphi +b{\tan }^2\frac{\theta }{2}$

$\Rightarrow (a-b)(1-\cos \varphi )={\tan }^2\frac{\theta }{2}(a+b)(1+\cos \varphi )\Rightarrow {\tan }^2\frac{\theta }{2}=\frac{a-b}{a+b}.\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}$

$\Rightarrow \tan \frac{\theta }{2}=\sqrt{\left(\frac{a-b}{a+b}\right)}.\tan \frac{\theta }{2}$