Answer
We know that, $(a+b)^{2}=a^{2}+b^{2}+2\times\,a\times\,b$
Hence from the question,
$(\sin\theta+\cos\theta)^{2}=(\sin\theta)^{2}+(\cos\theta)^{2}+2\times\,(\sin\theta)\times\,(\cos\theta)
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1+2\sin\theta\cos\theta$
Since, $(\sin\theta)^{2}+(\cos\theta)^{2}=1$.
Hence proved.
Work Step by Step
We know that, $(a+b)^{2}=a^{2}+b^{2}+2\times\,a\times\,b$
Hence from the question,
$(\sin\theta+\cos\theta)^{2}=(\sin\theta)^{2}+(\cos\theta)^{2}+2\times\,(\sin\theta)\times\,(\cos\theta)
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1+2\sin\theta\cos\theta$
Since, $(\sin\theta)^{2}+(\cos\theta)^{2}=1$.
Hence proved.