Answer
$1 - 2\sin\theta\cos\theta $
Work Step by Step
Given expression-
$ (\sin\theta-\cos\theta)^{2} $
= $ (\sin\theta-\cos\theta) (\sin\theta-\cos\theta) $
( recall $a^{2} = a\times a$)
= $ \cos^{2}\theta -\cos\theta\sin\theta -\sin\theta\cos\theta + \sin^{2}\theta$
(By FOIL method)
= $ \cos^{2}\theta + \sin^{2}\theta - 2\sin\theta\cos\theta $
= $1 - 2\sin\theta\cos\theta $
( From first Pythagorean identity $ \cos^{2}\theta + \sin^{2}\theta$ = 1 )
