prove thst sin square theta + cos square theta = 1 Share with your friends Share 1 Manbar Singh answered this 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 4 View Full Answer Shubha answered this Sin O = p/h and Cos O = b/hTherefore Sin2O + Cos2O = p2/h2 +b2/h2or Sin2O + Cos2O = (p2+b2)/h2But according to pythagoran theorem p2 + b2 = h2Therefore sin2O + Cos2O = h2/h2 = 1 (Prooved) 1