Question

Prove the following that:

`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ

Answer 1

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