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Question
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
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Solution
LHS = `tan^3 θ/(1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ)`
= `tan^3 θ/sec^2 θ + cot^3 θ/(cosec^2 θ)`
= 1 + tan2θ = sec2θ; 1 + cot2θ = cosec2θ
= `sin^3 θ/cos^3 θ xx cos^2 θ + cos^3 θ/sin^3 θ xx sin^2 θ`
= `sin^3 θ/cos θ + cos^3 θ/sin θ`
= `(sin^4 θ + cos^4 θ)/(cos θ.sin θ)`
= `((sin^2θ)^2 + (cos^2θ)^2)/(sin θ.cos θ)`
= `((sin^2 θ + cos^2 θ)^2 - 2 sin^2 θ. cos^2 θ)/(sin θ.cos θ)` ...[a2 + b2 = (a + b)2 − 2ab]
= `((1)^2 - 2sin^2θ. cos^2θ)/(sin θ.cos θ)`
= `(1 - 2sin^2θ. cos^2θ)/(sinθ.cosθ)`
= `1/(sinθ.cosθ) - (2sin^2θ. cos^2θ)/(sinθ.cosθ)`
= secθ. cosecθ − 2 sinθ cosθ
= RHS
Hence proved.
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