# Write the other trigonometric ratios of A in terms of sec A.

## Solution :

Consider a triangle ABC, in which $$\angle$$ B = 90

For $$\angle$$ A, we have :

Base = AB,

Perp = BC

and Hyp = AC

$$\therefore$$    sec A = $$Hyp\over Base$$ = $$AC\over AB$$

So,  $$AC\over AB$$  = sec A = $$sec A\over 1$$

Let AB = k and AC = k sec A

So,  BC = $$\sqrt{{AC}^2 – {AB}^2}$$ = $$k\sqrt{sec^2 A – 1}$$

Now, sin A = $$BC\over AB$$ = $$\sqrt{sec^2 A – 1}\over sec A$$

cos A = $$AB\over AC$$ = $$1\over sec A$$

tan A = $$BC\over AB$$ = $$\sqrt{sec^2 A – 1}$$

cot A = $$1\over tan A$$ = $$1\over \sqrt{sec^2 A – 1}$$

cosec A = $$1\over sin A$$ = $$sec A\over \sqrt{sec^2 A – 1}$$