# Prove that 1 + $$tan^2 \theta$$ = $$sec^2 \theta$$.

## Solution :

In right angled triangle ABC,

$$sec \theta$$ = $$AC\over AB$$  $$\implies$$   $$sec^2 \theta$$ = $$AC^2\over AB^2$$

$$tan \theta$$ = $$BC\over AB$$  $$\implies$$   $$tan^2 \theta$$ = $$BC^2\over AB^2$$

$$\implies$$ 1 + $$tan^2 \theta$$ = 1 + $$BC^2\over AB^2$$  = $$AB^2 + BC^2\over AB^2$$ = $$AC^2\over AB^2$$

[ By Pythagoras theorem,  $$AC^2$$ = $$BC^2 + AB^2$$ ]

$$\implies$$ 1 + $$tan^2 \theta$$ = $$sec^2 \theta$$