Question

Question #6a1a5

Answer 1

`(sin2x)/(1-2cos2x)`

`=(sin2x)/(sin^2x+cos^2x-2(cos^2x-sin^2x))`

`=(sin2x)/(sin^2x+cos^2x-2cos^2x+2sin^2x)`

`=(2sinxcosx)/(3sin^2x-cos^2x)`

`=((2sinxcosx)/cos^2x)/((3sin^2x)/cos^2x-cos^2x/cos^2x)`

`=(2tanx)/(3tan^2x-1)`