Question

How sin square theta plus cos square theta equal 1?

Answer 1

here is your answer

Answer 2

     

$\sf{We know that,} \\ \\ \sin\theta=\frac{\sf{opposite\ side}}{\sf{hypotenuse}} \\ \\ \sf{\&} \\ \\ \cos\theta=\frac{\sf{adjacent\ side}}{\sf{hypotenuse}}$

$\sf{Let the opposite side, adjacent side and hypotenuse be} \ a,\ b\ \sf{and} \ c\ \sf{respectively.}$

$\sf{So that} \ \ a^2+b^2=c^2$

$\sin\theta=\frac{a}{c} \ \ \ \ \ \Longrightarrow\ \ \ \ \ \sin^2\theta=(\frac{a}{c})^2=\frac{a^2}{c^2} \\ \\ \\ \cos\theta=\frac{b}{c} \ \ \ \ \ \Longrightarrow\ \ \ \ \ \cos^2\theta=(\frac{b}{c})^2=\frac{b^2}{c^2} \\ \\ \\ \\ \boxed{\sin^2\theta+\cos^2\theta=\frac{a^2}{c^2}+\frac{b^2}{c^2}=\frac{a^2+b^2}{c^2}=\frac{c^2}{c^2}=1}$

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

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