Prove that:

Question:
Prove that:

$\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$

Solution:

To Prove: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$

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

Proof:

$\mathrm{LHS}=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}$

$=\tan ^{-1} \frac{1}{2}+\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\left(\frac{1}{5} \times \frac{1}{8}\right)}\right)$

$=\tan ^{-1} \frac{1}{2}+\tan ^{-1}\left(\frac{8+5}{40-1}\right)$

$=\tan ^{-1} \frac{1}{2}+\tan ^{-1}\left(\frac{13}{39}\right)$

$=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$

$=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\left(\frac{1}{2} \times \frac{1}{3}\right)}\right)$

$=\tan ^{-1}\left(\frac{3+2}{6-1}\right)$

$=\tan ^{-1} 1$

$=\frac{\pi}{4}$

$=\mathrm{RHS}$

Therefore $\mathrm{LHS}=\mathrm{RHS}$

Hence proved.