Question

The value of $x$ satisfying the equation tan${}^{-1}$ x + tan${}^{-1}$$\left(\frac{2}{3}\right)$ tan${}^{-1}$ $\left(\frac{7}{4}\right)$is equal to

• A. $\frac{1}{2}$
• B. -$\frac{1}{2}$
• C. $\frac{3}{2}$
• D. -$\frac{1}{3}$
• E. $\frac{1}{3}$

Answer 1

Answer: (A)

We have,
${\tan }^{-1}x+{\tan }^{-1}\left(\frac{2}{3}\right)={\tan }^{-1}\left(\frac{7}{4}\right)$
${\tan }^{-1}\left[\frac{x+\frac{2}{3}}{1-\frac{2}{3}(x)}\right]={\tan }^{-1}\left(\frac{7}{4}\right)\Rightarrow \frac{\frac{3x+2}{3}}{\frac{3-2x}{3}}=\frac{7}{4}$
$\Rightarrow 4(3x+2)=7(3-2x)$
$\Rightarrow 12x+8=21-14x$
$\Rightarrow 26x=13\Rightarrow x=\frac{13}{26}=\frac{1}{2}$