Find the value of $\frac{\sqrt{1+\sin 2A}+\sqrt{1-\sin 2A}}{\sqrt{1+\sin 2A}-\sqrt{1-\sin 2A}}$ when $\left|\tan A\right|<1$ and $\left|A\right|$ is acute

The correct option is C. $\cot A$

Given expression is $\frac{\sqrt{1+\sin 2A}+\sqrt{1-\sin 2A}}{\sqrt{1+\sin 2A}-\sqrt{1-\sin 2A}}$

$=\frac{\sqrt{{\sin }^{2}A+{\cos }^{2}A+\sin 2A}+\sqrt{{\sin }^{2}A+{\cos }^{2}A-\sin 2A}}{\sqrt{1+\sin 2A}-\sqrt{1-\sin 2A}}$

$=\frac{\sqrt{{(\cos A+\sin A)}^2}+\sqrt{{(\cos A-\sin A)}^2}}{\sqrt{{(\cos A+\sin A)}^2}-\sqrt{{(\cos A-\sin A)}^2}}$

$=\frac{|\cos A+\sin A|+|\cos A-\sin A|}{|\cos A+\sin A|-|\cos A-\sin A|}$

$=\frac{\cos A+\sin A+\cos A-\sin A}{\cos A+\sin A-(\cos A-\sin A)}$

$(\because -\frac{\pi }{4}<A<\frac{\pi }{4}$ and in this interval $\cos A>\sin A)$

$=\frac{2\cos A}{2\sin A}=\cot A$ when $\left|\tan A\right|<1$ and $\left|A\right|$ is acute.