If $\cot \theta = \frac{7}{8},$ evaluate:
$(i) \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii) \cot^2 \theta$

(N/A) Let us consider a right triangle $ABC,$ right-angled at point $B.$
$\cot \theta = \frac{\text{Side adjacent to } \angle \theta}{\text{Side opposite to } \angle \theta} = \frac{BC}{AB} = \frac{7}{8}.$
If $BC = 7k,$ then $AB = 8k,$ where $k$ is a positive integer.
Applying the Pythagoras theorem in $\triangle ABC,$
$AC^2 = AB^2 + BC^2 = (8k)^2 + (7k)^2 = 64k^2 + 49k^2 = 113k^2.$
$AC = \sqrt{113}k.$
Now,$\sin \theta = \frac{AB}{AC} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}$ and $\cos \theta = \frac{BC}{AC} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}.$
$(i) \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} = \frac{1-\sin^2 \theta}{1-\cos^2 \theta} = \frac{1-(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2} = \frac{1-\frac{64}{113}}{1-\frac{49}{113}} = \frac{\frac{49}{113}}{\frac{64}{113}} = \frac{49}{64}.$
$(ii) \cot^2 \theta = (\cot \theta)^2 = (\frac{7}{8})^2 = \frac{49}{64}.$