Question

If $\cot \theta =\frac{7}{8}$, Then $\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}+\frac{15}{64}=$

Answer 1

$cot\theta =\frac{7}{8}=\frac{B}{P}$ (Refer to Image)

$\Rightarrow H=\sqrt{7^2+8^2}=\sqrt{49+64}=\sqrt{113}$

$\sin \theta =\frac{P}{H}=\frac{8}{\sqrt{113}}$

$cos\theta =\frac{B}{H}=\frac{7}{\sqrt{113}}$

Now, putting the value in expression,

$\Rightarrow \frac{(1+\frac{8}{\sqrt{113}})(1-\frac{8}{\sqrt{113}})}{(1-\frac{7}{\sqrt{113}})(1-\frac{7}{\sqrt{113}})}+\frac{15}{64}$

$\Rightarrow \frac{(1-\frac{64}{113})}{(1-\frac{49}{113})}+\frac{15}{64}$

$\Rightarrow \frac{113-64}{113-49}+\frac{15}{64}$

$\Rightarrow \frac{49}{64}+\frac{15}{64}=\frac{64}{64}=1$