Question

${tan}^{- 1} \frac{1}{2} + {tan}^{- 1} \frac{1}{4} =___________$

\begin{array}{l} \begin{array}{rcl} Since we know that \\ {tan}^{- 1} A + {tan}^{- 1} B & = & {tan}^{- 1} (\frac{A + B}{1 - A \cdot B}) \end{array} \\ Therefore according to this formula, we have \\ \begin{array}{rcl} A = \frac{1}{2} & and & B = \frac{1}{4} \\ So, {tan}^{- 1} \frac{1}{2} + {tan}^{- 1} \frac{1}{4} & = & {tan}^{- 1} (\frac{\frac{1}{2} + \frac{1}{4}}{1 - \frac{1}{2} \cdot \frac{1}{4}}) \\ = & {tan}^{- 1} (\frac{\frac{3}{4}}{\frac{7}{8}}) \\ = & {tan}^{- 1} (\frac{6}{7}) \end{array} \end{array}

So, correct answer is option-2, ${tan}^{- 1} \frac{6}{7}$