Question

The answer is D but can I know the step solution?

Answer 1

Explanation:

Apparently, you're looking for the particular form of the identity that applies in the given angle range.

You can expand the "1" under the radical, then take the square root of the resulting perfect square:

$\sqrt{1-2\sin{\theta}\cos{\theta}}=\sqrt{\sin^2{\theta}+\cos^2{\theta}-2\sin{\theta}\cos{\theta}}\\\\=\sqrt{(\sin{\theta}-\cos{\theta})^2}=|\sin{\theta}-\cos{\theta}|=\pm(\sin{\theta}-\cos{\theta})$

In the 4th quadrant, sine is negative and cosine is positive, so the difference will be positive when sine is subtracted from cosine:

$=\cos{\theta}-\sin{\theta} \qquad\text{in the 4th quadrant}$