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If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2

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Question

If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2 

Theorem
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Solution

a cosθ + b sinθ = m  ......(i)

a sinθ - b cosθ = n  ......(ii)

Squaring and adding equations 1 and 2, we get,

(a cosθ + b sinθ)2 + (a sinθ - b cosθ)2 = m2 + n2

⇒ a2cos2θ + b2sin2θ + 2ab sin θ cos θ + a2sin2θ + b2cos2θ - 2ab sin θ cos θ = m2 + n2

⇒ a2cos2θ + b2sin2θ + a2sin2θ + b2cos2θ = m2 + n2

⇒ a2(sin2θ + cos2θ) + b2(sin2θ + cos2θ) = m2 + n2

Using, sin2θ + cos2θ = 1

We get,

⇒ a2 + b2 = m2 + n2

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Chapter 3: Trigonometric Functions - Exercise [Page 53]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 7 | Page 53

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