Answer
$1 - 2\cos\theta\sin\theta $
Work Step by Step
Given expression-
$ (\cos\theta-\sin\theta)^{2} $
= $ \cos^{2}\theta -2\cos\theta\sin\theta + \sin^{2}\theta$
(Recall $(a+b)^{2} = a^{2} -2ab + b^{2}$)
= $ \cos^{2}\theta + \sin^{2}\theta - 2\cos\theta\sin\theta $
= $1 - 2\cos\theta\sin\theta $
( From first Pythagorean identity $ \cos^{2}\theta + \sin^{2}\theta$ = 1 )
