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Question
Prove that:
`(sin^3 theta + cos^3 theta)/(sin theta + cos theta) = 1 - sin theta cos theta`
Solution
LHS = `(sin^3 theta + cos^3 theta)/(sin theta + cos theta)`
`= ((sin theta + cos theta)(sin^2 theta - sin theta cos theta + cos^2 theta))/(sin theta + cos theta)` ...[∵ a3 + b3 = (a + b)(a2 - ab + b2)]
= sin2 θ - sin θ cos θ + cos2 θ
= 1 - sin θ cos θ ...[sin2 θ + cos2 θ = 1]
= RHS
Hence Proved.
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