Question

If ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)-{\tan }^{-1}\left(\frac{x}{3}\right)=0$ then $x$ is equal to

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

Answer 1

The correct option is D

$\frac{13}{3}$

Explanation for the correct option

Given that, ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)-{\tan }^{-1}\left(\frac{x}{3}\right)=0$

$\begin{array}{crclc}\Rightarrow & {\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)& =& {\tan }^{-1}\left(\frac{x}{3}\right)& \\ \Rightarrow & {\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{3}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{4}}}\right)& =& {\tan }^{-1}\left(\frac{x}{3}\right)& \left[\because {\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{{\displaystyle x+y}}{1-{\displaystyle x}y}\right)\right]\\ \Rightarrow & {\tan }^{-1}\left(\frac{{\displaystyle 13}}{9}\right)& =& {\tan }^{-1}\left(\frac{x}{3}\right)& \\ \Rightarrow & \frac{13}{9}& =& \frac{x}{3}& \\ \Rightarrow & x& =& \frac{13}{3}& \end{array}$

Hence the correct option is option(D) i.e. $\frac{13}{3}$