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.`

Answer 1

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.