Question

What is `tan^2theta/(1-sin^3theta` in terms of non-exponential trigonometric functions?

Answer 1

`tan^2theta/(1-sin^3theta)`

`=sin^2theta/(cos^2theta(1-sin^3theta))`

`=(8sin^2theta)/(8cos^2theta-8cos^2thetasin^3theta)`

`=(8sin^2theta)/(8cos^2theta-2xx(2costhetasintheta)^2sintheta`

`=(8sin^2theta)/(8cos^2theta-2sin^2(2theta)sintheta`

`=(4(1-cos2theta))/(4(1+cos2theta)-(1-cos(4theta))sintheta)`