If $A={\tan }^{-1}\left(\frac{x\sqrt{3}}{2k-x}\right)$ and $B={\tan }^{-1}\left(\frac{2x-k}{k\sqrt{3}}\right)$, then the value of $A-B$ in degrees, is

We find that

 $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{\frac{x\sqrt{3}}{2k-x}-\frac{2x-k}{k\sqrt{3}}}{1+\frac{x\sqrt{3}}{2k-x}\cdot \frac{2x-k}{k\sqrt{3}}}=\frac{1}{\sqrt{3}}$

$\Rightarrow A-B={30}^{\circ }$