If θ is an acute angle such that tan2 θ=87, then the value of (1+sin θ) (1−sin θ )(1+cos θ) (1−cos θ)is

Given that: $\tan ^{2} \theta=\frac{8}{7}$ and $\theta$ is an acute angle

We have to find the following expression

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

Since

$\tan ^{2} \theta=\frac{8}{7}$

$\tan \theta=\sqrt{\frac{8}{7}}$

$\tan \theta=\frac{\sqrt{8}}{\sqrt{7}}$

Since $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

$\Rightarrow$ Perpendicular $=\sqrt{8}$

$\Rightarrow$ Base $=\sqrt{7}$

$\Rightarrow$ Hypotenuse $=\sqrt{8+7}$

$\Rightarrow$ Hypotenuse $=\sqrt{15}$

We know that $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ and $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

We find:

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

$=\frac{\left(1+\frac{\sqrt{8}}{\sqrt{15}}\right)\left(1-\frac{\sqrt{8}}{\sqrt{15}}\right)}{\left(1+\frac{\sqrt{7}}{\sqrt{15}}\right)\left(1-\frac{\sqrt{7}}{\sqrt{15}}\right)}$

$=\frac{\left(1-\frac{8}{15}\right)}{\left(1-\frac{7}{15}\right)}$

$=\frac{\frac{7}{15}}{\frac{8}{15}}$

$=\frac{7}{8}$

Hence the correct option is $(a)$