Answer
(a). $a^{2} -2ab + b^{2} $
(b). $1 - 2\cos\theta\sin\theta $
Work Step by Step
(a). $ (a-b)^{2} $
= $ (a-b) (a-b) $
= $ a^{2} -ab -ba + b^{2}$
= $ a^{2} -2ab + b^{2}$
(b). $ (\cos\theta-\sin\theta)^{2} $
= $ (\cos\theta-\sin\theta) (\cos\theta-\sin\theta) $
= $ \cos^{2}\theta -\cos\theta\sin\theta -\sin\theta\cos\theta + \sin^{2}\theta$
= $ \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 )
