The Solution:
Given the expression below:
[tex]\frac{\left(sin\theta+cos\theta\right)^2}{1+2sin\theta\:cos\theta}[/tex]
We are required to simplify the above expression.
[tex]\begin{gathered} (\sin \theta+\cos \theta)^2=\sin ^2\theta+2\sin \theta\cos \theta+\cos ^2\theta=\sin ^2\theta+\cos ^2\theta+2\sin \theta\cos \theta \\ =1+2\sin \theta\cos \theta \\ \text{ Since }\sin ^2\theta+\cos ^2\theta=1 \end{gathered}[/tex]
So,
[tex]\frac{(sin\theta+cos\theta)^2}{1+2sin\theta\: cos\theta}=\frac{1+2sin\theta\: cos\theta}{1+2sin\theta\: cos\theta}=1[/tex]
Therefore, the correct answer is:
[tex]\frac{(sin\theta+cos\theta)^2}{1+2sin\theta\: cos\theta}=1[/tex]
