Prove that:

To Prove: $\tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\frac{\pi}{2}$

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

Proof:

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

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

$=\tan ^{-1} 1+\tan ^{-1}\left(\frac{5}{6-1}\right)$

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

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

$=\frac{\pi}{2}$

$=\mathrm{RHS}$

Therefore LHS = RHS

Hence proved.