Question

Solve for x the following :

(a) ${tan}^{-1}\frac{x-1}{x-2}+{tan}^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$

(b) ${tan}^{-1}\frac{x-1}{x+1}+{tan}^{-1}\frac{2x-1}{2x+1}={tan}^{-1}\frac{23}{36}$

Answer 1

(a)

$={\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)$

$={\tan }^{-1}\left(\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2}\times \frac{x+1}{x+2}}\right)$

$={\tan }^{-1}\left(\frac{2x^2-4}{-3}\right)$

$=\frac{\pi }{4}$

$\left(\frac{2x^2-4}{-3}\right)={\tan }^{-1}\frac{\pi }{4}$

$\left(\frac{2x^2-4}{-3}\right)=1$

$2x^2-4=-3$

$x^2=\frac{1}{2}$

$\Rightarrow x=\frac{1}{\sqrt{2}}$

(b)

$={\tan }^{-1}\left(\frac{x-1}{x+1}\right)+{\tan }^{-1}\left(\frac{2x-1}{2x+1}\right)$

$={\tan }^{-1}\left(\frac{\frac{x-1}{x+1}+\frac{2x-1}{2x+1}}{1-\frac{x-1}{x+1}\times \frac{2x-1}{2x+1}}\right)$

$={\tan }^{-1}\left(\frac{4x^2-2}{6x}\right)$

$={\tan }^{-1}\frac{23}{36}$

$\left(\frac{4x^2-2}{6x}\right)=\frac{23}{36}$

$6(4x^2-2)=23x$

$24x^2-23x-12=0$

Solving the quadratic equation, we get,

$x=\frac{4}{3}$ and $x=\frac{-3}{8}$