Show that $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}$

(A) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
Given $L.H.S. = \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{2}{11}$.
Applying the formula with $x = \frac{1}{2}$ and $y = \frac{2}{11}$:
$L.H.S. = \tan ^{-1} \left( \frac{\frac{1}{2} + \frac{2}{11}}{1 - (\frac{1}{2} \times \frac{2}{11})} \right)$.
Simplifying the numerator: $\frac{1}{2} + \frac{2}{11} = \frac{11 + 4}{22} = \frac{15}{22}$.
Simplifying the denominator: $1 - \frac{2}{22} = 1 - \frac{1}{11} = \frac{10}{11}$.
Thus,$L.H.S. = \tan ^{-1} \left( \frac{15/22}{10/11} \right) = \tan ^{-1} \left( \frac{15}{22} \times \frac{11}{10} \right)$.
$L.H.S. = \tan ^{-1} \left( \frac{15}{2 \times 10} \right) = \tan ^{-1} \left( \frac{15}{20} \right) = \tan ^{-1} \frac{3}{4}$.
Since $L.H.S. = R.H.S.$,the identity is proved.