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Question
Prove the following:
`(sin^3theta + cos^3theta)/(sintheta + costheta) + (sin^3theta - cos^3theta)/(sintheta - costheta)` = 2
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Solution
L.H.S. = `(sin^3theta + cos^3theta)/(sintheta + costheta) + (sin^3theta - cos^3theta)/(sintheta - costheta)`
`=((sintheta + costheta)(sin^2theta - sintheta costheta + cos^2theta))/(sintheta + costheta) + ((sintheta - costheta)(sin^2theta + sintheta costheta + cos^2theta))/(sintheta - costheta)` `...[a^3+b^3=(a+b)(a^2-ab+b^2),a^3-b^3=(a-b)(a^2+ab+b^2)]`
= (sin2θ + cos2θ – sinθ cosθ) + (sin2θ + cos2θ + sinθ cosθ)
= 2 (sin2θ + cos2θ)
= 2(1)
= 2
= R.H.S.
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