Question

If $\tan \theta =\frac{8}{15}$, evaluate $\frac{(2+2\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(2-2\cos \theta )}$.

Answer 1

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

$\Rightarrow \sin \theta =\frac{8}{17}$ & $\cos \theta =\frac{15}{17}$

$\therefore \frac{(2+2\times \left(\frac{8}{17}\right))(1-\frac{8}{17})}{(1+\frac{15}{17})(2-2\times \left(\frac{15}{17}\right))}$

$=\frac{(2+\frac{16}{17})\left(\frac{17-8}{17}\right)}{\left(\frac{17+15}{17}\right)\left(\frac{34-30}{17}\right)}$

$=\frac{\frac{50}{17}\times \frac{9}{17}}{\frac{32}{17}\times \frac{4}{17}}$

$=\frac{{}^{25}\text{/}50\times 9}{32\times {\text{/}4}_2}=\frac{225}{64}$.