Question

$\text{Solve}:{\tan }^{-1}4x+{\tan }^{-1}6x=\frac{\pi }{4}$

Answer 1

${\tan }^{-1}4x+{\tan }^{-1}6x=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-1}\left(\frac{4x+6x}{1-24x^2}\right)=\frac{\pi }{4},\text{provided}4x\times 6x<1\Rightarrow \frac{4x+6x}{1-24x^2}={\tan }^{-1}\left(\frac{\pi }{4}\right)\Rightarrow \frac{10x}{1-24x^2}=1\Rightarrow (12x-1)(2x+1)=0\Rightarrow x=\frac{1}{12},\frac{-1}{2}$

$\text{For}x=\frac{1}{12},4x\times 6x=\frac{1}{6}<1$
$\text{For}x=\frac{-1}{2},4x\times 6x=6\nless 1$
Therefore, the solution of the given equation is $x=\frac{1}{12}$