Q5. Given $$sec\theta \ = \ \frac{13}{12}$$ , calculate all other trigonometric ratios.

Given, $$sec\theta \ = \ \frac{13}{12} \ = \ \frac{H}{B} \ = \ \frac{AC}{AB}$$

Let AC = 13k and AB = 12k.

Then by Pythagoras theorem,

$$BC \ = \ \sqrt{AC^2 - AB^2} \ = \ \sqrt{(13k)^2 - (12k)^2} \ = \ \sqrt{169k^2 - 144k^2} \ = \ \sqrt{25k^2} \ = \ 5k$$

∴ $$sin\theta \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{5k}{13k} \ = \ \frac{5}{13}$$

$$cos\theta \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{12k}{13k} \ = \ \frac{12}{13}$$

$$tan\theta \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{5}{12}$$

$$cot\theta \ = \ \frac{1}{tan\theta} \ = \ \frac{12}{5}$$

$$cosec\theta \ = \ \frac{1}{sin\theta} \ = \ \frac{13}{5}$$