The value of ( 1 - ) ( + ) ( + ) ( 1 + ) + ( 1 + ) is equal to:

Answer: Option D

$$\begin{array}{rl}& \frac{\sec \theta \left(1-\sin \theta \right)\left(\sin \theta +\cos \theta \right)\left(\sec \theta +\tan \theta \right)}{\sin \theta \left(1+\tan \theta \right)+\cos \theta \left(1+\cot \theta \right)}\\ & =\frac{\left(\sec \theta -\tan \theta \right)\left(\sin \theta +\cos \theta \right)\left(\sec \theta +\tan \theta \right)}{\sin \theta \left(1+\tan \theta \right)+\cos \theta \left(\frac{1+\tan \theta }{\tan \theta }\right)}\\ & =\frac{\left({\sec }^2\theta -{\tan }^2\theta \right)\left(\sin \theta +\cos \theta \right)}{\left(1+\tan \theta \right)\left(\sin \theta +\frac{{\cos }^2\theta }{\sin \theta }\right)}\\ & =\frac{1\left(\sin \theta +\cos \theta \right)}{\left(1+\tan \theta \right)\left(\frac{1}{\sin \theta }\right)}\\ & =\frac{\left(\sin \theta +\cos \theta \right)}{\left(\cos \theta +\sin \theta \right)}\sin \theta .\cos \theta \\ & =\sin \theta .\cos \theta \end{array}$$