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

Solving LHS, $\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}$

Let $\tan^{-1}\frac{1}{4}=x$

$\Rightarrow \tan x=\frac{1}{4}$

Squaring both sides,

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

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

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

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

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

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

$Since,\: \sin^{2}x=1-\cos^{2}x$

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

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

Again,

Let $\tan^{-1}\frac{2}{9}=y$

$\Rightarrow \tan y=\frac{2}{9}$

Squaring both sides,

$\Rightarrow \tan^{2}y=\frac{4}{81}$

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

$\Rightarrow \sec^{2}y=\frac{85}{81}$

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

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

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

Since, $\sin^{2}y=1-\cos^{2}y$

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

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

We know that, $\sin(x+y)=\sin x.\sin y+ \cos x.\sin y$

$\Rightarrow \sin\left ( x+y \right )=\frac{1}{\sqrt{17}}.\frac{9}{\sqrt{85}}+ \frac{4}{\sqrt{17}}.\frac{2}{\sqrt{85}}$

$\Rightarrow \sin\left ( x+y \right )=\frac{17}{\sqrt{17}.\sqrt{85}}$

$\Rightarrow \sin\left ( x+y \right )=\frac{\sqrt{17}}{\sqrt{17}.\sqrt{5}}$

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

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

Since , LHS=RHS

Hence Proved