Question

If ${\text{tan}}^{-1}\left(\frac{x-1}{x+1}\right)+{\text{tan}}^{-1}\left(\frac{2x-1}{2x+1}\right)={\text{tan}}^{-1}\frac{23}{36},$ then

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

Answer 1

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

Let ${\text{tan}}^{-1}\left(\frac{x-1}{x+1}\right)=A$ and ${\text{tan}}^{-1}\left(\frac{2x-1}{2x+1}\right)=B$
Then $\tan A=\frac{x-1}{x+1}$ and $\tan B=\frac{2x-1}{2x+1}$

$A+B={\tan }^{-1}\frac{23}{36}$
$\Rightarrow \tan (A+B)=\frac{23}{36}$
$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{23}{36}$
$\Rightarrow \frac{\frac{x-1}{x+1}+\frac{2x-1}{2x+1}}{1-\frac{x-1}{x+1}\times \frac{2x-1}{2x+1}}=\frac{23}{36}$
$\Rightarrow \frac{2x^2-1}{3x}=\frac{23}{36}$
$\Rightarrow 24x^2-23x-12=0\Rightarrow 24x^2-32x+9x-12=0$
$\Rightarrow (3x-4)(8x+3)=0\Rightarrow x=\frac{4}{3},-\frac{3}{8}$

But for $x=-\frac{3}{8},$ both terms of LHS are negative and hence, LHS is negative, while RHS is positive.
So, the only solution is $x=\frac{4}{3}$