Answer
0
Work Step by Step
Given expression
= $( \sin 45^{\circ} - \cos 45^{\circ})^{2}$
= $ \sin^{2} 45^{\circ}$ - 2 $ \sin 45^{\circ} \cos 45^{\circ}$ + $ \cos^{2} 45^{\circ}$
[ on expanding using identity $(a-b)^{2} = a^{2} - 2ab + b^{2}$]
= $(\frac{1}{\sqrt 2})^{2}$ - (2$ \times \frac{1}{\sqrt 2} \times\frac{1}{\sqrt 2})$ + $ (\frac{1}{\sqrt 2})^{2}$
( substituting for exact values of trigonometric functions)
=$\frac{1}{2} - (2\times \frac{1}{2}) + \frac{1}{2}$
= $\frac{1}{2} + \frac{1}{2} - (2\times \frac{1}{2}) $
= 1 - 1
= 0
