Correct Answer: $\sqrt{\frac{1-\sin 2 A}{\cos 2 A}}$

Solution : Given expression,
$\sqrt{\frac{1-\tan A}{1+\tan A}}$
= $\sqrt{\frac{1-\frac{\sin A}{\cos A}}{1+\frac{\sin A}{\cos A}}}$
= $\sqrt{\frac{\cos A-\sin A}{\cos A+\sin A}}$
Multiplying and dividing by $\sqrt{\cos A-\sin A}$
$\sqrt{\frac{\cos A-\sin A}{\cos A+\sin A}}\times\sqrt{\frac{\cos A-\sin A}{\cos A-\sin }}$
= $\sqrt{\frac{(\cos A-\sin A)^2}{(\cos A+\sin A)(\cos A-\sin A)}}$
= $\sqrt{\frac{\sin^2A+\cos^2A-2\sin A \cos A}{\cos^2 A-\sin^2 A}}$
We know, $\sin^2\theta+\cos^2\theta=1,\ \sin2\theta=2\sin\theta\cos\theta\text{ and }\cos2\theta=\cos^2 \theta-\sin^2 \theta$
= $\sqrt{\frac{1-\sin2A}{\cos2A}}$
Hence, the correct answer is $\sqrt{\frac{1-\sin2A}{\cos2A}}$.