( ^3 ^2 + ^3 cose c^2 + 2 ) ÷ (1 + cosec2θ + tan2θ), 0° < θ < 90°, is equal to:

Answer: Option C

$$\begin{array}{rl}& \left(\frac{{\tan }^3\theta }{{\sec }^2\theta }+\frac{{\cot }^3\theta }{\text{cose}{\text{c}}^2\theta }+2\sin \theta \cos \theta \right)÷\left(1+\text{cose}{\text{c}}^2\theta +{\tan }^2\theta \right)\\ & =\left(\frac{{\sin }^3\theta }{\cos \theta }+\frac{{\cos }^3\theta }{\text{sin}\theta }+2\sin \theta \cos \theta \right)÷\left(\text{cose}{\text{c}}^2\theta +{\sec }^2\theta \right)\\ & =\left(\frac{{\sin }^4\theta +{\cos }^4\theta +2{\sin }^2\theta {\cos }^2\theta }{\sin \theta \cos \theta }\right)÷\left(\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }\right)\\ & =\left(\frac{{\left({\sin }^2\theta +{\cos }^2\theta \right)}^2}{\sin \theta \cos \theta }\right)÷\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }\right)\\ & =\left(\frac{1}{\sin \theta \cos \theta }\right)\times \left(\frac{{\sin }^2\theta {\cos }^2\theta }{1}\right)\\ & =\sin \theta \cos \theta \end{array}$$