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Question
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
Solution
LHS: `((sin^3θ)/(cos^3θ))/((1 + sin^2θ)/(cos^2θ)) + ((cos^3θ)/(sin^3θ))/((1 + cos^2θ)/(sin^2θ))`
= `((sin^3θ)/(cos^3θ))/(((cos^2θ + sin^2θ))/cos^2θ) + ((cos^3θ)/(sin^3θ))/(((sin^2θ + cos^2θ))/sin^2θ)`
= `sin^3θ/cosθ + cos^3θ/sinθ`
= `(sin^4θ + cos^4θ)/(cosθsinθ)`
= `((sin^2θ + cos^2θ)^2 - 2 sin^2θ cos^2θ)/(cosθ sinθ)`
= `(1 - 2 sin^2θ cos^2θ)/(cosθ sinθ)`
= `1/(cos θ sinθ) - (2 sin^2θcos^2θ)/(cosθ sinθ)`
= secθ cosec θ – 2 sinθ cosθ
= RHS
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