Question

Prove that : $\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}]+\tan [\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}]=\frac{2b}{a}$

Answer 1

Consider the L.H.S.

$=\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}]+\tan [\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}]$

Let $\theta =\frac{1}{2}{\cos }^{-1}\frac{a}{b}$ …….. (1)

Therefore,

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

We know that

$\tan (A+B)=\frac{\tan A+\mathrm{tanB}}{1-\tan A\tan B}$

Therefore,

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

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

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

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

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

We know that

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

Therefore,

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

From equation (1), we get

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

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

Hence, proved