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

Answer :

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