# Prove that $$sin^2 \theta$$ + $$cos^2 \theta$$ = 1.

## Solution :

In right angled triangle ABC,

$$sin \theta$$ = $$BC\over AC$$  $$\implies$$   $$sin^2 \theta$$ = $$BC^2\over AC^2$$

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

On adding,

$$sin^2 \theta$$ + $$cos^2 \theta$$ = $$BC^2\over AC^2$$ + $$AB^2\over AC^2$$

$$sin^2 \theta$$ + $$cos^2 \theta$$ = $$BC^2 + AB^2\over AC^2$$ = $$AC^2\over AC^2$$ = 1

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

$$\implies$$ $$sin^2 \theta$$ + $$cos^2 \theta$$ = 1