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

30°

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

$⇒A-B=\tan^{-1}(\frac{1}{\sqrt{3}})=30°$