# If a cos θ + b sinθ = c, then prove that a sinθ – b cos θ = ± Let

(a cos θ + b sin θ)2 + (a sin θ – b cos θ)2 = a2cos2θ + b2 sin2θ + 2ab cos θ sin θ + a2sin2θ

+ b2 cos2θ – 2ab cos θ sin θ

c2 + (a sin θ – b cos θ)2 = a2 (cos2 θ + sin2 θ) + b2 (cos2 θ + sin2 θ)

c2 + (a sin θ – b cos θ)2 = a2 + b2

(a sin θ – b cos θ)2 = a2 + b2 – c2

(a sin θ – b cos θ) = ±√ (a2 + b2 – c2)

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