Question

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

Answer 1

L.H.S

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

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

$\cos 2\theta =\frac{a}{b}$ ……. (2)

Therefore,

$=\tan (\frac{\pi }{4}+\theta )+{\tan }^{-1}(\frac{\pi }{4}-\theta )$

$=\frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\tan \theta }+\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta }$

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

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

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

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

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

$=\frac{2}{\cos 2\theta }$

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

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

Hence, proved.