Question

Prove: ${\sin }^2\theta +{\cos }^2\theta =1$

Answer 1

Let a, b, c be lengths of right angled triangle

By definition

$\sin \theta =b/c\left(\frac{oppositeside}{hypotenuse}\right)$

$\cos \theta =a/c\left(\frac{adjacentside}{hypotenuse}\right)$

${\sin }^2\theta +{\cos }^2\theta =\frac{b^2}{c^2}+\frac{a^2}{c^2}=\frac{a^2+b^2}{c^2}$

From Pythagoras theorem

$c^2=a^2+b^2$

$\therefore \frac{a^2+b^2}{c^2}=1$

${\sin }^2\theta +{\cos }^2\theta =1$

Hence, proved.