Question

Prove that , $tan(\frac{\pi }{2}+\frac{1}{2}\arccos \frac{a}{b})+tan(\frac{\pi }{4}-\frac{1}{2}\arccos ab)=\frac{2b}{a}$

Answer 1

$\tan (\frac{\pi }{2}+\frac{1}{2}arc\cos \frac{a}{b}+\tan (\frac{\pi }{4}-\frac{1}{2}arc\cos ab)$

Let $arc\cos \frac{a}{b}=\theta \Rightarrow \cos \theta =\frac{a}{b}$

$\tan (\frac{\pi }{4}+\frac{\theta }{2})+\tan (\frac{\pi }{4}-\frac{\theta }{2})$

$=\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}+\frac{1-\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}}$

$=\frac{(1+\tan \frac{\theta }{2}{)}^2+(1-\tan \frac{\theta }{2}{)}^2}{1-{\tan }^2\frac{\theta }{2}}$

$=\frac{2(1+{\tan }^2\frac{\theta }{2})}{1-{\tan }^2\frac{\theta }{2}}$

$=\frac{2}{\cos \theta }(\therefore \cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta })$

$=\frac{2b}{a}$