Answer
See the explanation below.
Work Step by Step
${{\left( \sin \theta -\cos \theta \right)}^{2}}=1-\sin 2\theta $
Recall the algebraic formula.
${{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy$
Consider the left side of the given expression and apply the algebraic formula.
${{\left( \sin \theta -\cos \theta \right)}^{2}}={{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta $
Recall the Trigonometric Identities.
$\begin{align}
& \sin 2\theta =2\sin \theta \cos \theta \\
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
\end{align}$
Apply the Trigonometric Identities.
${{\left( \sin \theta -\cos \theta \right)}^{2}}=1-\sin 2\theta $
Hence, it is proved that the given identity ${{\left( \sin \theta -\cos \theta \right)}^{2}}=1-\sin 2\theta $ holds true.
