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)}

Ans:  

\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}

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