# Given $$sec \theta$$ = $$13\over 12$$, calculate all other trigonometric ratios.

## Solution :

Consider a triangle ABC in which $$\angle$$ A = $$\theta$$ and $$\angle$$ B = 90

Then, Base = AB, perp = BC and Hypo. = AC,

$$\therefore$$  $$sec \theta$$ = $$perp\over hypo$$ = $$BC\over AC$$ = $$3\over 4$$

Let AC = 13k and AB = 12k. Then,

By using Pythagoras Theorem,

$${BC}^2$$ = $${AC}^2$$ – $${AB}^2$$

= $$169k^2$$ – $$125k^2$$ = $$25k^2$$

$$\implies$$ BC = 5k

$$sin \theta$$ = $$BC\over AC$$ = $$5\over 13$$

$$cos \theta$$ = $$AB\over AC$$ = $$12\over 13$$

$$tan \theta$$ = $$BC\over AB$$ = $$5\over 12$$

$$cot \theta$$ = $$AB\over BC$$ = $$12\over 5$$

$$cosec \theta$$ = $$AC\over BC$$ = $$13\over 5$$