Question

Prove that $\tan \left\{\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right\}+\tan \left\{\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right\}=\frac{2b}{a}$

Answer 1

Let $\frac{1}{2}{\text{cos}}^{-1}\frac{a}{b}=\theta ⟹\cos 2\theta =\frac{a}{b}$

$LHS=\tan (\frac{\pi }{4}+\theta )+\tan (\frac{\pi }{4}-\theta )=\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }=\frac{(1+\tan \theta {)}^2+(1-\tan \theta {)}^2}{1-{\tan }^2\theta }=\frac{2({\sin }^2\theta +{\cos }^2\theta )}{{\cos }^2\theta -{\sin }^2\theta }=\frac{2}{\cos 2\theta }=\frac{2b}{a}$