Prove that

To Prove: $(\sin x-\cos x)^{2}=1-\sin 2 x$

Taking LHS,

$=(\sin x-\cos x)^{2}$

Using,

$(a-b)^{2}=\left(a^{2}+b^{2}-2 a b\right)$

$=\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x$

$=\left(\sin ^{2} x+\cos ^{2} x\right)-2 \sin x \cos x$

$=1-2 \sin x \cos x\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$=1-\sin 2 x[\because \sin 2 x=2 \sin x \cos x]$

= RHS

∴ LHS = RHS

Hence Proved