Question

If cot θ = , then $\frac{(2+2\mathrm{sin\theta })(1-\mathrm{sin\theta })}{(1+\mathrm{cos\theta })(2-2\mathrm{cos\theta })}$ = ?
(a) $\frac{64}{225}$
(b) $\frac{225}{64}$
(c) $\frac{64}{289}$
(d) $\frac{289}{64}$

Answer 1

(b) $\frac{225}{64}$
We have:

$$\begin{gathered} \frac{\left(2+2\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(2-2\cos \theta \right)} \\ =\frac{2\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{2\left(1+\cos \theta \right)\left(1-\cos \theta \right)} \\ =\frac{2\left(1-{\sin }^2\theta \right)}{2\left(1-{\cos }^2\theta \right)} \\ =\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta =\frac{225}{64} \end{gathered}$$