If $\cot \theta=\frac{15}{8}$ then evaluate $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
Given : $\cot \theta=\frac{15}{8}$
Since, $\cot \theta=\frac{B}{P}$
$\Rightarrow P=8$ and $B=15$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow 8^{2}+15^{2}=H^{2}$
$\Rightarrow H^{2}=64+225$
$\Rightarrow H^{2}=289$
$\Rightarrow H=17$
Therefore,
$\sin \theta=\frac{P}{H}=\frac{8}{17}$
$\cos \theta=\frac{B}{H}=\frac{15}{17}$
Now,
$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{1-\sin ^{2} \theta}{1-\cos ^{2} \theta}$
$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta} \quad\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$
$=\cot ^{2} \theta$
$=\left(\frac{15}{8}\right)^{2}$
$=\frac{225}{64}$
Hence, $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{225}{64}$.
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