Question

If xcosθ – ysinθ = a, xsinθ + ycos θ = b, prove that x²+y²=a²+b².

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Answer 1

$a = xcos \theta - ysin \theta \\ b = xsin \theta + ycos \theta \\ \\ {a}^{2} + {b}^{2} = ( {xcos \theta - ysin \theta})^{2} + ( {xsin \theta + ycos \theta})^{2} \\ \\ {a}^{2} + {b}^{2} = {x}^{2} {cos}^{2} \theta + {y}^{2} {sin}^{2} \theta - 2xycos \theta \: sin \theta + {x}^{2} {sin}^{2} \theta + {y}^{2} {cos}^{2} \theta + 2xysin \theta \: cos \theta \\ \\ {a}^{2} + {b}^{2} = {x}^{2} ( {cos}^{2} \theta + {sin}^{2} \theta) + {y}^{2} ( {sin}^{2} \theta + {cos}^{2} \theta) \\ \\ {a}^{2} + {b}^{2} = {x}^{2} + {y}^{2}$