Question

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

Answer 1

The given expression is:

$\frac{(2+2\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(2-2\cos \theta )}$

$=\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$

$=\frac{1-{\sin }^2\theta }{1-{\cos }^2\theta }$

$=\frac{{\cos }^2\theta }{{\sin }^2\theta }$

$={\cot }^2\theta$

Putting $\cot \theta =\frac{15}{8}$, we get

${\cot }^2\theta ={\left(\frac{15}{8}\right)}^2=\frac{225}{64}$