Question

If ${\tan }^{-1}\left(2x\right)+{\tan }^{-1}\left(3x\right)=\pi /4$ then $x=$ $?$

Answer 1

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

We know that,

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

Using the above identity we get

${\tan }^{-1}2x+{\tan }^{-1}3x={\tan }^{-1}\left(\frac{2x+3x}{1-2x\cdot 3x}\right)={\tan }^{-1}\left(\frac{5x}{1-6x^2}\right)$

$\therefore {\tan }^{-1}\frac{5x}{1-6x^2}=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\frac{5x}{1-6x^2}={\tan }^{-1}1$

$\Rightarrow \frac{5x}{1-6x^2}=1$

$\Rightarrow -6x^2+1=5x$

$\Rightarrow 6x^2+5x-1=0$

$\Rightarrow 6x^2+6x-x-1=0$

$\Rightarrow 6x(x+1)-1(x+1)=0$

$\Rightarrow (6x-1)(x+1)=0$

$\therefore x=\frac{1}{6}or-1$