Prove that sinA-cosA square=1-sin2A
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LHS
= (sin A - cos A)^2
= sin^2 A + cos^2 A - 2 sin A cos A
= 1 - 2 sin A cos A
because sin^2 x + cos^2 x = 1
so,
= 1 - 2sinAcosA
= 1- sin2A because sin 2A = 2sinAcosA
= 1 - sin 2A = RHS
= (sin A - cos A)^2
= sin^2 A + cos^2 A - 2 sin A cos A
= 1 - 2 sin A cos A
because sin^2 x + cos^2 x = 1
so,
= 1 - 2sinAcosA
= 1- sin2A because sin 2A = 2sinAcosA
= 1 - sin 2A = RHS
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Answer:
LHS=
sin²A + cos²A - 2sinAcosA
1-2sinAcosA
1-2[sin2A/2]
⇒1-sin2A (because sinAcosA = sin2A/2)
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