Question

If $\begin{array}{rcl}{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)-{\tan }^{-1}\left(\frac{x}{3}\right)& =& 0\end{array}$, then $x$ is equal

• A. $\frac{7}{3}$
• B. $3$
• C. $\frac{11}{3}$
• D. $\frac{13}{3}$

Answer 1

The correct option is D

$\frac{13}{3}$

Solve the equation to find $x$:

$\begin{array}{rcl}{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)-{\tan }^{-1}\left(\frac{x}{3}\right)& =& 0\\ {\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)& =& {\tan }^{-1}\left(\frac{x}{3}\right)\\ {\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{3}{4}}{1-\left(\frac{1}{3}\right)\left(\frac{3}{4}\right)}\right)& =& {\tan }^{-1}\left(\frac{x}{3}\right)\\ \frac{x}{3}& =& \frac{13}{9}\\ x& =& \frac{13}{3}\end{array}$

Hence option (D) is the answer.