Q) If cot θ = $\frac{15}{8}$ then evaluate that $\frac{(2+2~sin~\theta)(1-sin~\theta)}{(1+cos~\theta)(2-2~cos~\theta)}$

$\frac{(2+2~sin~\theta)(1-sin~\theta)}{(1+cos~\theta)(2-2~cos~\theta)}$ $=\frac{2(1+sin~\theta)(1-sin~\theta)}{2(1+cos~\theta)(1-cos~\theta)}$ $=\frac{\cancel{2}(1-sin^{2}\theta)}{\cancel{2}(1-cos^{2}\theta)}$ $=\frac{cos^{2}\theta}{sin^{2}\theta}$ $=(\frac{cos~\theta}{sin~\theta})^{2}$ $=(cot~\theta)^{2}$ $\because~cot~\theta = \frac{15}{8}$ $\implies \frac{(2+2~sin~\theta)(1-sin~\theta)}{(1+cos~\theta)(2-2~cos~\theta)}$ $=(cot~\theta)^{2}=(\frac{15}{8})^{2}=\frac{225}{64}$ Please do press “Heart” button if you liked the solution.