Question

If ${\tan }^{-1}(x+1)+{\tan }^{-1}(x-1)-{\tan }^{-1}\frac{8}{31}$=0,then x is $\frac{1}{4}$.If true enter 1 else 0.

Answer 1

$ta{n}^{-1}(x+1)+ta{n}^{-1}(x-1)-ta{n}^{-1}\frac{8}{31}=0$$

$\Rightarrow ta{n}^{-1}(x+1)+ta{n}^{-1}(x-1)=ta{n}^{-1}\frac{8}{31}$

$\Rightarrow ta{n}^{-1}\left[\frac{x+1+1-1}{1-(x+1)(x-1)}\right]=ta{n}^{-1}\frac{8}{31}$

$\Rightarrow ta{n}^{-1}\left[\frac{2x}{1-(x^2-1)}\right]=ta{n}^{-1}\frac{8}{31}$

$\Rightarrow ta{n}^{-1}\left[\frac{2x}{2-x^2}\right]=ta{n}^{-1}\frac{8}{31}$

$\Rightarrow \frac{2x}{2-x^2}=\frac{8}{31}$

$\Rightarrow 62x=16-8x^2$

$\Rightarrow 8x^2+62x-16=0$

$\Rightarrow 4x^2+31x-8=0$

$\Rightarrow 4x^2+32x-x-8=0$

$\Rightarrow 4x(x+8)-1(x+8)=0$

$\Rightarrow 4x-1=0,x+8=0$

$x=\frac{1}{4},x=-8$ not possible

S0, $x=\frac{1}{4}$