Question

If $\cot \theta =\frac{15}{8}$, then $\frac{(2+2\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(2-2\cos \theta )}=$

• A. $\frac{64}{225}$
• B. $\frac{225}{64}$
• C. $\frac{64}{289}$
• D. $\frac{289}{64}$

Answer 1

Answer: (B)

$\frac{225}{64}$

$\frac{(2+2\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(2-2\cos \theta )}=\frac{2(1+\sin \theta )(1-\sin \theta )}{2(1+\cos \theta )(1-\cos \theta )}=\frac{1-{\sin }^2\theta }{1-{\cos }^2\theta }$
$=\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta ={\left(\frac{15}{8}\right)}^2=\frac{225}{64}$.