Question

If ${\tan }^{-1}\left(\frac{x-1}{x+1}\right)+{\tan }^{-1}\left(\frac{2x-1}{2x+1}\right)={\tan }^{-1}\frac{23}{36};x>1,$ then the value of $x$ can be

• A. $\frac{4}{3}\text{or}-\frac{3}{8}$
• B. $-\frac{3}{8}$
• C. $\frac{4}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{4}{3}$

Given : ${\tan }^{-1}\left(\frac{x-1}{x+1}\right)+{\tan }^{-1}\left(\frac{2x-1}{2x+1}\right)={\tan }^{-1}\frac{23}{36}$
We have $x>1$
Now,
$\frac{x-1}{x+1}=1-\frac{2}{x+1}\Rightarrow \frac{x-1}{x+1}\in (0,1)\frac{2x-1}{2x+1}=1-\frac{2}{2x+1}\Rightarrow \frac{2x-1}{2x+1}\in (\frac{1}{3},1)\therefore \frac{x-1}{x+1}\times \frac{2x-1}{2x+1}\in (0,1)$

Now, the equation becomes
$\frac{\frac{x-1}{x+1}+\frac{2x-1}{2x+1}}{1-\frac{x-1}{x+1}\times \frac{2x-1}{2x+1}}=\frac{23}{36}\Rightarrow \frac{2x^2-1}{3x}=\frac{23}{36}\Rightarrow 24x^2-23x-12=0\Rightarrow 24x^2-32x+9x-12=0\Rightarrow (3x-4)(8x+3)=0\Rightarrow x=\frac{4}{3}(\because x>1)$

Hence, the only solution is $x=\frac{4}{3}$