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Question

acosθ+bsinθ=m and asinθbcosθ=n, then a2+b2

A
m2n2
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B
m2n2
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C
n2m2
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D
m2+n2
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Solution

The correct option is D m2+n2
Given: a cosθ+b sinθ=m and a sinθb cosθ=nm2+n2=(a cosθ+b sinθ)2+(a sinθb cosθ)2m2+n2=a2cos2θ+b2sin2θ+2ab cosθ sinθ+a2sin2θ+b2cos2θ2ab sinθcosθm2+n2=a2(cos2θ+sin2θ)+b2(sin2θ+cos2θ)m2+n2=a2+b2 (sin2θ+cos2θ=1)

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