Correct Answer: 0

Solution : $a =\cot A+\cos A$
$b =\cot A-\cos A$
We have to find the value of $a^2-b^2-4 \sqrt{ab}$.
$= (\cot A+\cos A)^{2}-(\cot A-\cos A)^{2}-4\sqrt{(\cot A+\cos A)(\cot A-\cos A)}$
$= \cot^{2} A+\cos^{2} A+2\cot A\times \cos A-(\cot^{2} A+\cos^{2} A-2\cot A\times \cos A)-4\sqrt{\cot^{2}A-\cos^{2}A}$
$= (\cot^{2} A+\cos^{2} A+2\cot A\times \cos A-\cot^{2} A-\cos^{2} A+2\cot A\times \cos A)-4\sqrt{\frac{\cos^{2}A}{\sin^{2}A}-\cos^{2}A}$
$= (4\cot A\cos A)-4\sqrt{\frac{\cos^{2}A(1-\sin^{2}A)}{\sin^{2}A}}$
$= (4\cot A\cos A)-4\sqrt{\cos^{2}A\times \frac{\cos^{2}A}{\sin^{2}A}}$
$= (4\cot A\cos A)-4\sqrt{\cos^{2}A\times \cot^{2}A}$
$= (4\cot A\cos A)-4\cos A\times \cot A$
$= 0$
Hence, the correct answer is 0.