Question

If cot$\theta$ = $\frac{7}{8}$, then the value of $\frac{(1+\cos \theta )}{(1-\sin \theta )}.\frac{(1-\cos \theta )}{(1+\sin \theta )}$ is:

• A. $\frac{49}{64}$
• B. $\frac{8}{7}$
• C. $\frac{64}{49}$
• D. $\frac{7}{8}$

Answer 1

Answer: (C)

$\frac{64}{49}$

$\frac{(1+\cos \theta )(1-\cos \theta )}{(1-\sin \theta )(1+\sin \theta )}=\frac{1-{\cos }^2\theta }{1-{\sin }^2\theta }=\frac{{\sin }^2\theta }{{\cos }^2\theta }={\tan }^2\theta =\frac{1}{{\cot }^2\theta }=(\frac{8}{7}{)}^2=\frac{64}{49}$