Question

The value of $\tan \left[{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)\right]$ is

• A. $\frac{5}{6}$
• B. $\frac{7}{6}$
• C. $\frac{1}{6}$
• D. $\frac{1}{7}$

Answer 1

The correct option is D

$\frac{1}{7}$

Explanation for the correct option

The given trigonometric expression: $\tan \left[{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)\right]$.

It is known that ${\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{A}\mathbf{\right)}-{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{A}\mathbf{-}\mathbf{B}}{\mathbf{1}\mathbf{+}\mathbf{A}\mathbf{B}}\mathbf{\right)}$.

$$\begin{gathered} \Rightarrow \tan \left[{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)\right]=\tan \left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}}{1+\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{3}}\right)}\right)\right] \\ \Rightarrow \tan \left[{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)\right]=\frac{{\displaystyle \frac{\left(1\times 3\right)-\left(1\times 2\right)}{6}}}{1+{\displaystyle \frac{1}{6}}} \\ \Rightarrow \tan \left[{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)\right]=\frac{{\displaystyle \frac{1}{6}}}{{\displaystyle \frac{6+1}{6}}} \\ \Rightarrow \tan \left[{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)\right]=\frac{{\displaystyle \frac{1}{6}}}{{\displaystyle \frac{7}{6}}} \\ \Rightarrow \tan \left[{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)\right]=\frac{1}{7} \end{gathered}$$

Hence, option D is correct.