Answer
We know that $(a+b)\times(a-b)=(a^{2}-b^{2})$
Also, $(\sin\theta)^{2}+(\cos\theta)^{2}=1$
Now, based on the question, we have,
$(1+\sin\theta)\times(1-\sin\theta)=1-(\sin\theta)^{2}$
=$(\cos\theta)^{2}$
Hence proved.
Work Step by Step
We know that $(a+b)\times(a-b)=(a^{2}-b^{2})$
Also, $(\sin\theta)^{2}+(\cos\theta)^{2}=1$
Now, based on the question, we have,
$(1+\sin\theta)\times(1-\sin\theta)=1-(\sin\theta)^{2}$
=$(\cos\theta)^{2}$
Hence proved.