Q.1 Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Consider a $$∆ \ ABC$$

$$cotA \ = \ \frac{B}{P} \ = \ \frac{AB}{BC} \ => \ \frac{AB}{BC} \ = \ \frac{cotA}{1}$$

Let $$AB = kcotA$$ and $$BC = k$$.

By Pythagoras Theorem,

$$AC \ = \ \sqrt{AB^2 + BC^2} \ = \ \sqrt{k^2cot^2A + k^2} \ = \ k \sqrt{1 + cot^2A}$$

∴ $$sinA \ = \ \frac{P}{B} \ = \ \frac{BC}{AC} \ = \ \frac{k}{k \sqrt{1+cot^2A}} \ = \ \frac{1}{ \sqrt{1+cot^2A}}$$

$$secA \ = \ \frac{H}{B} \ = \ \frac{AC}{AB} \ = \ \frac{k \sqrt{1+cot^2A}}{kcotA} \ = \ \frac{ \sqrt{1+cot^2A}}{cotA}$$

and, $$tanA \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{k}{kcotA} \ = \ \frac{1}{cotA}$$