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Question
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Options
m2 − n2
m2n2
n2 − m2
m2 + n2
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Solution
Given:
`a cosθ+b sinθ= m,`
`a sinθ-b cos θ=n`
Squaring and adding these equations, we have
`(a cos θ+bsin θ)^2+(a sinθ-b cosθ)^2=(m)^2+(n)^2`
`⇒ (a^2 cos^2θ+b^2sin^2θ+2.a cosθ.bsinθ)+(a^2 sin^2θ+b^2 cos^2θ-2.a sin θ.bcosθ)=m^2+n^2`
`⇒ a^2 cos^2θ+b^2 sin^2θ+2ab sin θ cosθ+a^2 sin^2θ+b^2 cos^2θ-2ab sinθ cos θ=m^2+n^2`
`⇒a^2 cos^2θ+b^2 sin^2θ+a^2 sin^2θ+b^2 cos^2=m^2+n^2`
`⇒(a^2 cos^2θ+a^2 sin^2 θ)+(b^2 sin^2θ+b^2 cos^2θ)=m^2+n^2`
`⇒a^2 (cos^2θ+sin^2θ)+b^2(sin^2 θ+cos^2θ)=m^2+n^2`
`⇒ a^2(1)+b^2(1)=m^2+n^2`
`⇒ a^2+b^2=m^2+n^2`
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