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

Given, \( cot\theta \ = \ \frac{B}{P} \ = \ \frac{AB}{BC} \ = \ \frac{7}{8} \)

Let AB = 7k and BC = 8k.
Then by Pythagoras theorem,

\( AC \ = \ \sqrt{BC^2 + AB^2} \ \)
\( = \ \sqrt{(8k)^2 + (7k)^2} \ \)
\( = \ \sqrt{64k^2 + 49k^2} \ \)
\( = \ \sqrt{113k^2} \ = \ \sqrt{113}k \)



\(\therefore \) \( sin\theta \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{8k}{ \sqrt{113}k} \ = \ \frac{8}{\sqrt{113}} \)

and, \( cos\theta \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{7k}{\sqrt{113}k} \ = \ \frac{7}{\sqrt{113}} \)

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

= \( \frac{1 - \frac{64}{113}}{1- \frac{49}{113}} \)

= \( \frac{113 - 64}{113 - 49} \)

= \( \frac{49}{64} \)

(ii) \( cot^2\theta \ = \ ( \frac{7}{8} )^2 \ = \ \frac{49}{64} \)