Solve the equation.
${\cot }^{-1}x-{\cot }^{-1}(x+2)=15^{\circ }$

$=co{t}^{-1}x-co{t}^{-1}(x+2)=15^{\circ }$

$=co{t}^{-1}x=ta{n}^{-1}\frac{1}{x}$

$=ta{n}^{-1}\frac{1}{x}-ta{n}^{-1}\frac{1}{x+2}=15^{\circ }$

$=ta{n}^{-1}\frac{\frac{1}{x}-\frac{1}{x+2}}{1+\frac{1}{x(x+2)}}=15^{\circ }$

$=ta{n}^{-1}\frac{\frac{x+2-x}{x(x+2)}}{\frac{x^2+2x+1}{x(x+2)}}=15^{\circ }$

 $=ta{n}^{-1}\frac{2}{(x+1{)}^2}=15^{\circ }$

 $=\frac{2}{(x+1{)}^2}=tan15^{\circ }$

 $=\frac{2}{(x+1{)}^2}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

 Rationalising

$=\frac{2}{(x+1{)}^2}=\frac{2}{(\sqrt{3}+1{)}^2}$

$=⟹(x+1{)}^2=(\sqrt{3}+1{)}^2$

$⟹x+1=\sqrt{3}+1$ OR $x+1=-\sqrt{3}-1$

$⟹x=\sqrt{3}$ $OR$ $x=-2-\sqrt{3}$