If a cosθ + b sinθ = m and a sinθ – b cosθ = n, prove that : (m2

1.	If a cosθ + b sinθ = m and a sinθ – b cosθ = n, prove that : (m2 + n2 ) = (a2 + b2).
Answer» Given that a cosθ + bsinθ = m. … (1) And a sinθ – b cosθ = n. … (2) Now, squaring equation (2) and (3), we get m² = (a cosθ + sinθ)² = a² co²θ + b² sin²θ + 2absinθcosθ. … (3) And n² = (a sinθ − b cosθ)² = a² sin² + b² cos² – 2absinθcosθ. … (4) Now, adding equations (3) & (4), we get m² + n² = a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ) + 2ab sinθcosθ – 2ab sinθcosθ ⇒ m²+ n² = a² + b² . (∵ sin²θ + cos²θ = 1) Hence Proved.

If a cosθ + b sinθ = m and a sinθ – b cosθ = n, prove that : (m2 + n2 ) = (a2 + b2).

Answer»

Given that a cosθ + bsinθ = m. … (1)

And a sinθ – b cosθ = n. … (2)

Now, squaring equation (2) and (3), we get

m² = (a cosθ + sinθ)²

= a² co²θ + b² sin²θ + 2absinθcosθ. … (3)

And n² = (a sinθ − b cosθ)² = a² sin² + b² cos² – 2absinθcosθ. … (4)

Now, adding equations (3) & (4), we get

m² + n² = a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ) + 2ab sinθcosθ – 2ab sinθcosθ

⇒ m²+ n² = a² + b² . (∵ sin²θ + cos²θ = 1)

Hence Proved.

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If a cosθ + b sinθ = m and a sinθ – b cosθ = n, prove that : (m2 + n2 ) = (a2 + b2).

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