The expression (tanθ + cotθ)(secθ + tanθ)(1 - sinθ), 0° < θ < 90° is equal to:

Answer: Option C

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