Question

Evaluate :
${\tan }^{-1}\left(\frac{4x}{1+5x^2}\right)+{\tan }^{-1}\left(\frac{2+3x}{3-2x}\right)$

Answer 1

${\tan }^{-1}\left(\frac{4}{1+5x^2}\right)+{\tan }^{-1}\left(\frac{2+3x}{3-2x}\right)$

we know that ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$

${\tan }^{-1}\left(\frac{4}{1+5x^2}\right)+{\tan }^{-1}\left(\frac{2+3x}{3-2x}\right)=\left(\frac{\frac{4}{1+5x^2}+\frac{2+3x}{3-2x}}{1-\left(\frac{4}{1+5x^2}\right)\left(\frac{2+3x}{3-2x}\right)}\right)$

$={\tan }^{-1}\left(\frac{4(3-2x)+(2+3x)(1+5x^2)}{(1+5x^2)(3-2x)-4(2+3x)}\right)$

$={\tan }^{-1}\left(\frac{12-8x+2+10x^2+3x+15x^3}{3-2x+15x^2-10x^3-8-12x}\right)$

$={\tan }^{-1}\left(\frac{15x^3+10x^3-5x+14}{-10x^3+15x^2-14x-5}\right)$