Question

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

• A. $90°$
• B. $60°$
• C. $45°$
• D. ${\tan }^{-1}2$

Answer 1

The correct option is C

$45°$

Explanation for the correct option:

Solve the given expression:

$2{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)$$={\tan }^{-1}\left(\frac{2\times {\displaystyle \frac{1}{3}}}{1-{\left({\displaystyle \frac{1}{3}}\right)}^2}\right)+\tan -1\left(\frac{1}{2}\right)$ $\left[\mathbf{\because }\mathbf{2}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathbf{=}\frac{\mathbf{2}\mathbf{A}}{\mathbf{1}\mathbf{-}{\mathbf{A}}^{\mathbf{2}}}\right]$

$=ta{n}^{-1}\left(\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{8}{9}}}\right)+ta{n}^{-1}\left(\frac{1}{2}\right)$

$={\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)$

$={\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{3}{4}}\times {\displaystyle \frac{1}{2}}}\right)$

$$\begin{gathered} ={\tan }^{-1}\left(\frac{{\displaystyle \frac{3+2}{8}}}{{\displaystyle \frac{8-3}{8}}}\right) \\ ={\tan }^{-1}\left(\frac{5}{5}\right) \end{gathered}$$

$$\begin{gathered} ={\tan }^{-1}\left(1\right) \\ =\mathbf{45}\mathbf{°} \end{gathered}$$

Hence, Option ‘C’ is Correct.