Question

If tanθ=78 , then the value of (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)=​

Answer 1

Given:

$tan\theta=\dfrac{7}{8}$

To Find:

The value of $(1+sin\theta)(1-sin\theta)(1+cos\theta)(1-cos\theta)$

Step-by-step explanation:

• The formula of $tan\theta$ is

$tan\theta=\dfrac{opposite\ side}{adjacent\ side}$

• Use Pythagoras theorem,

$(Hypotenuse)^{2}=(Opposite\ side)^{2}+(Adjacent\ side)^{2} \\(Hypotenuse)^{2}=7^{2}+8^{2}\\(Hypotenuse)^{2}=113\\Hypotenuse=\sqrt{113}$

$sin\theta=\dfrac{opposite\ side}{hypotenuse}=\dfrac{7}{\sqrt{113} }$

$cos\theta=\dfrac{Adjacent\ side}{hypotenuse} =\dfrac{8}{\sqrt{113} }$

• The value of

$=(1+sin\theta)(1-sin\theta)(1+cos\theta)(1-cos\theta)\\=(1-sin^{2} \theta)(1-cos^{2}\theta)$

$=(1-(\frac{7}{\sqrt{113} })^{2})(1-(\frac{8}{\sqrt{113} })^{2})\\=(1-\frac{49}{113})(1-\frac{64}{113})\\=(\frac{64}{113})(\frac{49}{113})\\=\dfrac{3136}{12769} .$