Question

Prove the following identities:

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

Answer 1

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.