Question

$\tan h^{-1}\left(\frac{1}{2}\right)+\tan h^{-1}\left(\frac{1}{3}\right)$ is equal to

• A. $\tan h^{-1}\left(\frac{5}{7}\right)$
• B. $\tan h^{-1}\left(\frac{7}{5}\right)$
• C. $\tan h^{-1}\left(\frac{1}{6}\right)$
• D. $\tan h^{-1}\left(\frac{5}{6}\right)$

Answer 1

The correct option is A

$\tan h^{-1}\left(\frac{5}{7}\right)$

Explanation for the correct option:

We know that $\tan h^{-1}\left(x\right)+\tan h^{-1}\left(y\right)=\tan h^{-1}\left(\frac{x+y}{1+xy}\right)$.

So,

$$\begin{gathered} \tan h^{-1}\left(\frac{1}{2}\right)+\tan h^{-1}\left(\frac{1}{3}\right)=\tan h^{-1}\left(\frac{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}}{1+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{3}}}\right) \\ =\tan h^{-1}\left(\frac{{\displaystyle \frac{5}{6}}}{{\displaystyle \frac{7}{6}}}\right) \\ =\tan h^{-1}\left(\frac{5}{7}\right) \end{gathered}$$

Hence, option (A) is the correct answer.