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Question
Prove the following identities:
`tan^3theta/(1 + tan^2theta) + cot^3theta/(1 + cot^2theta` = secθ cosecθ – 2sinθ cosθ
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Solution
L.H.S> = `tan^3theta/(1 + tan^2theta) + cot^3theta/(1 + cot^2theta`
= `tan^3theta/sec^2theta + cot^3theta/("cosec"^2theta)`
= `(((sin^3theta)/cos^3 theta))/((1/cos^2theta)) + (((cos^3 theta)/(sin^3 theta)))/((1/sin^2 theta)`
= `sin^3theta/costheta + cos^3theta/sintheta`
= `(sin^4theta + cos^4 theta)/(sintheta cos theta)`
= `((sin^2 theta)^2 + (cos^2 theta)^2)/(sin theta cos theta`
= `((sin^2 theta + cos^2 theta)^2 - 2sin^2 theta cos^2 theta)/(sintheta cos theta)` ...[∵ a2 + b2 = (a + b)2 - 2ab]
= `(1^2 - 2sin^2 theta cos^2 theta)/(sin theta cos theta)`
= `1/(costheta*sintheta) - (2sin^2theta cos^2 theta)/(sintheta cos theta)`
= secθ cosecθ – 2sinθ cosθ
= R.H.S.
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