In $\Delta ABC$, prove that: $$\tan^{-1}(\cot{B}\cot{C})+\tan^{-1}(\cot{C}\cot{A})+\tan^{-1}(\cot{A}\cot{B})=\tan^{-1}{\left(1+\frac{8\cos{A}\cos{B}\cos{C}}{\sin^{2}{2A} +\sin^{2}{2B} + \sin^{2}{2C}}\right)}.$$
Using $tan^{-1}x+\tan^{-1}y+tan^{-1}z =\tan^{-1}{\frac{x+y+z-xyz}{1+yz+zx+xy}}$ and a little simplication, I have arrived at $$\frac{\tan{A}\tan{B}\tan{C}-\cot{A}\cot{B}\cot{C}}{\tan{A}+\tan{B}+\tan{C} +\cot{A}+\cot{B}+\cot{C}}= 1+\frac{8\cos{A}\cos{B}\cos{C}}{\sin^{2}{2A} +\sin^{2}{2B} + \sin^{2}{2C}}.$$ How to proceed after here? Any kind of help/suggestions will be highly appreciated.