Question

Prove that : ${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}=\frac{1}{2}{\sin }^{-1}\frac{4}{5}$

Answer 1

L.H.S. $={\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}$
$={\tan }^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}.\frac{2}{9}}\right)$

$={\tan }^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right)={\tan }^{-1}\frac{1}{2}$ ..... (i)

R.H.S. $=\frac{1}{2}{\sin }^{-1}\frac{4}{5}=\alpha$ (say)

$2\alpha ={\sin }^{-1}\frac{4}{5}$

$\Rightarrow \sin 2\alpha =\frac{4}{5}$

$\Rightarrow \frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{4}{5}$

$\Rightarrow 4{\tan }^2\alpha -10\tan \alpha +4=0$

$\Rightarrow 4{\tan }^2\alpha -8\tan \alpha -2\tan \alpha +4=0$

$\Rightarrow 4\tan \alpha (\tan \alpha -2)-2(\tan \alpha -2)=0$

$\Rightarrow (\tan \alpha -2)(4\tan \alpha -2)=0$

$\tan \alpha =2,\tan \alpha =\frac{1}{2}$

$(\because -\frac{\pi }{2}\le 2\alpha \le \frac{\pi }{2})$

$\Rightarrow -\frac{\pi }{4}\le \alpha \le \frac{\pi }{4}$

$\therefore \tan \alpha =\frac{1}{2}\Rightarrow \alpha ={\tan }^{-1}\frac{1}{2}$ ....... (ii)

From (i) and (ii),
${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}=\frac{1}{2}{\sin }^{-1}\frac{4}{5}$