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

We have, ${\tan ^{ - 1}}\frac{1}{4} + {\tan ^{ - 1}}\frac{2}{9} = {\sin ^{ - 1}}\frac{1}{{\sqrt 5 }}$

….(i)
Let ${\tan ^{ - 1}}\frac{1}{4} = x$
$\Rightarrow$ $\tan x = \frac{1}{4}$

$\Rightarrow$ ${\tan ^2}x = \frac{1}{{16}}$
$\Rightarrow$ ${\sec ^2}x - 1 = \frac{1}{{16}}$

$\Rightarrow$ ${\sec ^2}x = 1 + \frac{1}{{16}} = \frac{{17}}{{16}}$

$\Rightarrow$ $\frac{1}{{{{\cos }^2}x}} = \frac{{17}}{{16}}$

$\Rightarrow$ ${\cos ^2}x = \frac{{16}}{{17}}$
$\Rightarrow$ $\cos x = \frac{4}{{\sqrt {17} }}$

$\Rightarrow$ ${\sin ^2}x = 1 - {\cos ^2}x = 1 - \frac{{16}}{{17}} = \frac{1}{{17}}$

$\Rightarrow$ $\sin x = \frac{1}{{\sqrt {17} }}$

…..(ii)
Again, let ${\tan ^{ - 1}}\frac{2}{9} = y$
$\Rightarrow$ $\tan y = \frac{2}{9} \Rightarrow {\tan ^2}y = \frac{4}{{81}}$

$\Rightarrow$ ${\sec ^2}y - 1 = \frac{4}{{81}}$
$\Rightarrow$ ${\sec ^2}y = \frac{4}{{81}} + 1 = \frac{{85}}{{81}}$

$\Rightarrow$ ${\cos ^2}y = \frac{{81}}{{85}} \Rightarrow \cos y = \frac{9}{{\sqrt {85} }}$

$\Rightarrow$ ${\sin ^2}y = 1 - {\cos ^2}y = 1 - \frac{{81}}{{85}} = \frac{4}{{85}}$

$\Rightarrow$ $\sin y = \frac{2}{{\sqrt {85} }}$

……(iii)
We know that,
$\sin (x + y) = \sin x \cdot \cos y + \cos x \cdot \sin y$

$= \frac{1}{{\sqrt {17} }} \cdot \frac{9}{{\sqrt {85} }} + \frac{4}{{\sqrt {17} }} \cdot \frac{2}{{\sqrt {85} }}$

$= \frac{{17}}{{\sqrt {17} \cdot \sqrt {85} }} = \frac{{\sqrt {17} }}{{\sqrt {17} \cdot \sqrt 5 }} = \frac{1}{{\sqrt 5 }}$

$\Rightarrow$ $(x + y) = {\sin ^{ - 1}}\frac{1}{{\sqrt 5 }}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{1}{4} + {\tan ^{ - 1}}\frac{2}{9} = {\sin ^{ - 1}}\frac{1}{{\sqrt 5 }}$ Hence proved.