If $\theta$ is an acute angle such that ${\text{tan}}^2\theta =\frac{8}{7}$, then the value of $\frac{(1+\text{sin}\theta )(1-\text{sin}\theta )}{(1+\text{cos}\theta )(1-\text{cos}\theta )}$ is

A. $\frac{7}{4}$

B. $\frac{7}{8}$

C. $\frac{8}{7}$

D. $\frac{64}{49}$

The correct option is B : $\frac{7}{8}$
$\frac{(1+\text{sin}\theta )(1-\text{sin}\theta )}{(1+\text{cos}\theta )(1-\text{cos}\theta )}$

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

$[\because {\text{cos}}^2\theta +{\text{sin}}^2\theta =1,so1-{\text{sin}}^2\theta ={\text{cos}}^2\theta ,1-{\text{cos}}^2\theta ={\text{sin}}^2\theta ]$

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

$=\frac{1}{{\text{tan}}^2\theta }[\because \frac{\text{sin}\theta }{\text{cos}\theta }=\text{tan}\theta ]$

$=\frac{7}{8}$