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