If a cos θ + b sin θ = m and a sin θ – b cos θ = n,

Question:

If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then show that a2 + b2 = m2 + n2.

Solution:

According to the question,

a cos θ + b sin θ = m …(i)

a sin θ – b cos θ = n …(ii)

Squaring and adding equation 1 and 2, we get,

(a cos θ + b sin θ)2 + (a sin θ – b cos θ)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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