Question

If $\cos \theta =-\frac{1}{2}$ and $\pi <\theta <\frac{3\pi }{2}$, then the value of $\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$ is

• A. $\frac{1}{3}$
• B. $\frac{3}{1}$
• C. $\frac{7}{4}$
• D. $\frac{64}{49}$

Answer 1

The correct option is A $\frac{1}{3}$

Given $\cos \theta =-\frac{1}{2}$
$\therefore \frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$
$=\frac{1-si{n}^2\theta }{1-{\cos }^2\theta }=\frac{{\cos }^2\theta }{si{n}^2\theta }=\frac{{\cos }^2\theta }{1-{\cos }^2\theta }$

$=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$