The value of ( 1 - 1 - )^2 + 1, 0° < θ < 90°, is equal to:

Answer: Option A

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