Question

If aSinA + bCosA = c
then prove that
aCosA - bSinA = √(a² + b² - c²)​

Answer 1

We have, asinθ+bcosθ=c

On squaring both sides, we get

(asinθ+bcosθ)2=c2

(a sin θ)2 + (b cos θ)2 + 2(a sin θ) (b cos θ) = c2

⇒ a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ = c2

⇒ a2(1 – cos2 θ) + b2 (1 – sin2 θ) + 2 ab sin θ cos θ = c2 [∵sin2θ+cos2θ=1]

⇒ a2 – a2 cos2 θ + b2 – b2 sin2 θ + 2ab sin θ cos θ = c2

⇒ –a2 cos2 θ – b2 sin2 θ + 2ab sin θ cos θ = c2 – a2 – b2

Taking Negative common,

⇒ a2 cos2 θ + b2 sin2 θ – 2ab sin θ cos θ = a2 + b2 – c2

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

⇒ (acosθ−bsinθ)2=a2+b2−c2

⇒acosθ−bsinθ = ±a2+b2−c2−−−−−−−−−−√

Hence proved, acosθ−bsinθ = a2+b2−c2−−

Answer 2

Answer:

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