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Question
Prove the following:
`((1 + cot theta + tan theta)(sin theta - costheta)) /(sec^3theta - "cosec"^3theta)`= sin2θ cos2θ
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Solution
L.H.S. = `((1 + cot theta + tan theta)(sin theta - costheta))/(sec^3theta - "cosec"^3theta)`
= `((1 + costheta/sintheta + sintheta/costheta)(sin theta - cos theta))/(1/cos^3theta - 1/sin^3theta)`
`=(((sintheta cos theta + cos^2theta + sin^2theta)/(sintheta costheta))(sintheta - cos theta))/((sin^3theta - cos^3theta)/(sin^3theta cos^3theta)`
`=((sintheta costheta + cos^2theta + sin^2theta)(sintheta - costheta))/(sintheta cos theta) xx (sin^3theta cos^3theta)/(sin^3theta - cos^3theta)`
`=((sintheta - costheta)(sin^2theta + sintheta costheta + cos^2theta)sin^2theta cos^2theta)/(sin^3theta - cos^3theta)`
`=((sintheta - costheta)(sin^2theta + sintheta costheta + cos^2theta)sin^2theta cos^2theta)/((sintheta - costheta)(sin^2theta + sintheta costheta + cos^2theta))` ...[∵ a3 – b3 = (a – b)(a2 + ab + b2)]
= sin2θ cos2θ
= R.H.S.
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