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Prove that (a sin theta + b cos theta )square + (a cos theta - b sin theta )square = a square + b squareWho will answer me correctly I will make it brainliest |
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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, +/-root a2+b2-c2=x Hence, acosθ - bsinθ = Root a^2 + b^2 - c^2. Hope U will FOUND it helpful.... Step-by-step explanation: |
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