Question
Prove that $\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}=\sin^{-1}\frac{1}{\sqrt{5}}.$
✓
Answer
We have, $\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}=\sin^{-1}\frac{1}{\sqrt{5}}$
LHS $=\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}$
$=\tan^{-1}\bigg(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times\frac{2}{9}}\bigg)$
$\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
$=\tan^{-1}\Big(\frac{17}{34}\Big)$
$=\tan^{-1}\Big(\frac{1}{2}\Big)$
$=\sin^{-1}\begin{pmatrix}\frac{\frac{1}{2}}{\sqrt{1+\big(\frac{1}{2}\big)^2}}\end{pmatrix}$
$\Big[\because\ \tan^{-1}\text{x}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)\Big]$
$=\sin^{-1}\Big(\frac{1}{\sqrt{5}}\Big)$
= RHS
Hence proved.
LHS $=\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}$
$=\tan^{-1}\bigg(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times\frac{2}{9}}\bigg)$
$\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
$=\tan^{-1}\Big(\frac{17}{34}\Big)$
$=\tan^{-1}\Big(\frac{1}{2}\Big)$
$=\sin^{-1}\begin{pmatrix}\frac{\frac{1}{2}}{\sqrt{1+\big(\frac{1}{2}\big)^2}}\end{pmatrix}$
$\Big[\because\ \tan^{-1}\text{x}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)\Big]$
$=\sin^{-1}\Big(\frac{1}{\sqrt{5}}\Big)$
= RHS
Hence proved.
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