Prove that:

Question:
Prove that:

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

Solution:

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

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

Proof:

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

$=\frac{\pi}{4}+\pi+\tan ^{-1}\left(\frac{2+3}{1-(2 \times 3)}\right)\{$ since $2 \times 3=6>1\}$

$=\frac{5 \pi}{4}+\tan ^{-1}\left(\frac{5}{-5}\right)$

$=\frac{5 \pi}{4}+\tan ^{-1}(-1)$

$=\frac{5 \pi}{4}-\frac{\pi}{4}$

$=\pi$

$=\mathrm{RHS}$

Therefore LHS = RHS

Hence proved.