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