Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. $-\sqrt{2}$

Answer 1

Answer: (A)

$1$

From Pythagoras theorem, we write

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

Dividing through by $c^2$ gives,

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

This can be simplified to:

${\left(\frac{a}{c}\right)}^2+{\left(\frac{b}{c}\right)}^2={\left(\frac{c}{c}\right)}^2$

Now, $\frac{a}{c}$ is $\frac{\text{opposite}}{\text{hypotenuse}}$, which is $\sin \theta$

$\frac{b}{c}$ is $\frac{\text{adjacent}}{\text{hypotenuse}}$, which is $\cos \theta$

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

So, option $A$ is correct.