Prove that

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, $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{225}{64}$.