The expression ^6 - ^6 + 3 ^2 \, ^2 ^2 + ^2 + 2 is equal to:

Answer: Option A

$$\begin{array}{rl}& \frac{{\tan }^6\theta -{\sec }^6\theta +3{\sec }^2\theta {\tan }^2\theta }{{\tan }^2\theta +{\cot }^2\theta +2}\\ & =\frac{{\left({\tan }^2\theta \right)}^3-{\left({\sec }^2\theta \right)}^3+3{\sec }^2\theta {\tan }^2\theta }{{\tan }^2\theta +{\cot }^2\theta +2}\\ & \left[a^3-b^3=\left(a-b\right)\left(a^2+b^2+ab\right)\right]\\ & =\frac{\left({\tan }^2\theta -{\sec }^2\theta \right)\left({\tan }^4\theta +{\sec }^4\theta +{\tan }^2\theta {\sec }^2\theta +3{\sec }^2\theta {\tan }^2\theta \right)}{{\tan }^2\theta +{\cot }^2\theta +2}\\ & =\frac{-1\left({\tan }^4\theta +{\sec }^4\theta +{\tan }^2\theta {\sec }^2\theta -3{\sec }^2\theta {\tan }^2\theta \right)}{{\tan }^2\theta +{\cot }^2\theta +2}\\ & =\frac{-1\left({\tan }^4\theta +{\sec }^4\theta -2{\sec }^2\theta {\tan }^2\theta \right)}{{\tan }^2\theta +{\cot }^2\theta +2}\\ & =\frac{-1\left({\sec }^2\theta -{\tan }^2\theta \right)}{{\tan }^2\theta +{\cot }^2\theta +2}\\ & =\frac{-1}{{\left(\tan \theta +\cot \theta \right)}^2}\\ & =\frac{-1}{{\left(\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }\right)}^2}\\ & =\frac{-1}{{\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }\right)}^2}\\ & =-{\left(\cos \theta \sin \theta \right)}^2\\ & =-{\cos }^2\theta {\sin }^2\theta \end{array}$$