Question

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If a sin thitta+ b cos 0(thitta) = c, then prove that a cos 0 (thitta)- b sin0 (thitta)= plus-minus [whole√a2 + b2 –c2].​

Answer 1

Answer:

Asinθ + bcosθ = c

taking square both sides,

(asinθ + bcosθ)² = c²

⇒a²sin²θ + b²cos²θ + 2absinθ.cosθ = c² --------(1)

Let acosθ - bsinθ = x

Squaring both sides

(acosθ - bsinθ)² = x²

⇒a²cos²θ + b²sin²θ -2absinθ.cosθ = x² ------(2)

Add equation (1) and (2),

a²sin²θ + b²cos²θ +2abinθ.cosθ + a²cos²θ + b²sin²θ -2absinθ.cosθ = c² + x²

⇒(a² + b²)cos²θ + (a² +b²)sin²θ = c² + x²

⇒(a² + b²)[sin²θ + cos²θ ] = c² + x²

⇒(a² + b²) = c² + x² [∵ sin²x + cos²x = 1 ]

⇒(a² + b² - c²) = x²

Take square root both sides,

\bold{\pm\sqrt{a^2 + b^2 - c^2} =x}

Hence, acosθ - bsinθ = \bold{\sqrt{a^2 + b^2 - c^2}}

hope this will help uhh