Prove that:

Solution:

To Prove: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \tan ^{-1} \frac{4}{3} \Rightarrow 2\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)=\tan ^{-1} \frac{4}{3}$

Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$

Proof:

$\mathrm{LHS}=2\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)$

$=2\left(\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4} \times \frac{2}{9}\right)}\right)\right)$

$=2 \tan ^{-1}\left(\frac{9+8}{36-2}\right)$

$=2 \tan ^{-1} \frac{17}{34}$

$=2 \tan ^{-1} \frac{1}{2}$

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

$=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{2}}{1-\left(\frac{1}{2} \times \frac{1}{2}\right)}\right)$

$=\tan ^{-1}\left(\frac{1}{\frac{4-1}{4}}\right)$

$=\tan ^{-1} \frac{4}{3}$

$=\mathrm{RHS}$

Therefore LHS $=$ RHS

Hence proved.