If cotθ = 15/8, then evaluate (2 + 2 sinθ)(1

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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