Question
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
Solution
R.H.S `m^2 sin^2 theta`
`= (a cos theta + b sin theta)^2 + (a sin theta - b cos theta)^2`
`= a^2 cos^2 theta + b^2 sin^2 theta + 2 ab sin theta cos theta + a^2 sin^2 theta + b^2 cos^2 theta - 2 ab sin theta cos theta`
`= a^2 cos^2 theta + b^2 cos^2 theta + b^2 sin^2 theta + a^2 sin^2 theta`
`= a^2(sin^2 theta + cos^2 theta) + b^2(sin^2 theta + cos^2 theta)`
`=a^2 + b^2` (∵ `sin^2 theta + cos^2 theta = 1`)
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Solution If a Cos θ + B Sin θ = M and a Sin θ – B Cos θ = N, Prove that A2 + B2 = M2 + N2 Concept: Trigonometric Identities.
