prove that sin square theta + cos square theta = 1
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Consider a right-angled triangle, ΔABC, where ∠BAC=θ,

By the Pythagorean theorem,AC2+BC2=AB2Dividing by AB2,⇒AC2AB2+BC2AB2=AB2AB2⇒(oppositehypotenuse)2+(adjacenthypotenuse)2=AB2AB2=1⇒sin2θ+cos2θ=1

By the Pythagorean theorem,AC2+BC2=AB2Dividing by AB2,⇒AC2AB2+BC2AB2=AB2AB2⇒(oppositehypotenuse)2+(adjacenthypotenuse)2=AB2AB2=1⇒sin2θ+cos2θ=1
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