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B
m2n2
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C
n2−m2
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D
m2+n2
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Solution
The correct option is Dm2+n2 Given:acosθ+bsinθ=mandasinθ−bcosθ=n⇒m2+n2=(acosθ+bsinθ)2+(asinθ−bcosθ)2⇒m2+n2=a2cos2θ+b2sin2θ+2abcosθsinθ+a2sin2θ+b2cos2θ−2absinθcosθ⇒m2+n2=a2(cos2θ+sin2θ)+b2(sin2θ+cos2θ)⇒m2+n2=a2+b2(∵sin2θ+cos2θ=1)