Question:
Prove that:
$\tan ^{-1} 2-\tan ^{-1} 1=\tan ^{-1} \frac{1}{3}$
Solution:
To Prove: $\tan ^{-1} 2-\tan ^{-1} 1=\tan ^{-1} \frac{1}{3}$
Formula Used: $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ where $x y>-1$
Proof:
$\mathrm{LHS}=\tan ^{-1} 2-\tan ^{-1} 1$
$=\tan ^{-1}\left(\frac{2-1}{1+2}\right)$
$=\tan ^{-1}\left(\frac{1}{3}\right)$
$=\mathrm{RHS}$
Therefore LHS = RHS
Hence proved.