Question

Solve the following equation
${\tan }^{-1}\frac{1}{1+2x}+{\tan }^{-1}\frac{1}{4x+1}={\tan }^{-1}\frac{2}{x^2}$.

Answer 1

${\tan }^{-1}\frac{1}{1+2x}+{\tan }^{-1}\frac{1}{4x+1}={\tan }^{-1}\frac{2}{x^2}$.

$[\because {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)]$

$\therefore {\tan }^{-1}\left[\frac{\frac{1}{1+2x}+\frac{1}{4x+1}}{1-\left(\frac{1}{1+2x}\right)\left(\frac{1}{4x+1}\right)}\right]={\tan }^{-1}\frac{2}{x^2}$

$\Rightarrow {\tan }^{-1}\left[\frac{\frac{4x+1+2x+1}{(1+2x)(1+4x)}}{\frac{(1+2x)(1+4x)-1}{(1+2x)(1+4x)}}\right]{\tan }^{-1}\frac{2}{x^2}$

$\Rightarrow {\tan }^{-1}\left(\frac{6x+2}{8x^2+6x}\right)=\Rightarrow {\tan }^{-1}\frac{2}{x^2}$

$\Rightarrow \frac{3x+1}{4x^2+3x}=\frac{2}{x^2}$

$\Rightarrow 3x^2+x^2=8x^2+6x$

$\Rightarrow x(3x^2-7x-6)=0$

$\Rightarrow x\left[3x\right(x-3)+(x-3\left)\right]=0$

$\Rightarrow x(x-3)(3x+2)=0$

$\Rightarrow x=0,x=3,x=-\frac{2}{3}$