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prove thst sin square theta + cos square theta = 1



Consider a right ∆ABC, in which ∠B = 90°.Let ∠ACB = θ    sin θ = perpendicularhypotenuse⇒ sin θ =ABAC⇒sin2θ = ABAC2⇒sin2θ =AB2AC2     ----1    cos θ = basehypotenuse⇒cos θ = BCAC⇒cos2θ = BCAC2⇒cos2θ = BC2AC2   -----2adding 1 and 2, we get   sin2θ + cos2θ =AB2AC2 +BC2AC2⇒ sin2θ + cos2θ =AB2 + BC2AC2    -----3In ∆ABC, AC2 = AB2 + BC2  Pythagoras Theorem    ------4Substituting the value of AC2 from 4 in 3, we get     sin2θ + cos2θ =AB2 + BC2AB2 + BC2⇒ sin2θ + cos2θ =1 

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Sin O = p/h and Cos O = b/h

Therefore Sin2O + Cos2O = p2/h2 +b2/h2

or Sin2O + Cos2O = (p2+b2)/h2

But according to pythagoran theorem p2 + b2 = h2

Therefore sin2O + Cos2O = h2/h2 = 1 (Prooved)

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