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


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

(a) $\frac{7}{8}$

(b) $\frac{8}{7}$

(c) $\frac{7}{4}$

(d) $\frac{64}{49}$


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

We have to find the following expression

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


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

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

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

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:

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





Hence the correct option is $(a)$

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