How do you simplify #(sinxcosx)/(1-sin^2x)#?

I got #cotx# as an answer... don't know if it is right though...

2 Answers
Nov 27, 2017

#(sinx*cosx)/(1-sin^2x)==tanx#

Explanation:

#(sinx*cosx)/(1-sin^2x)=(sinx*cosx)/cos^2x=sinx/cosx=tanx#

Nov 27, 2017

See the answer below...

Explanation:

#(sinxcosx)/(1-sin^2x)=(sinxcosx)/cos^2x#[BECAUSE #color(red)(sin^2x+cos^2x=1#]
#color(white)(mmmmfG)=(sinxcancelcosx)/(cosxcdotcancelcosx)=color(red)(tanx#

So the answer is #color(brown)(tanx#

Hope it helps...
Thank you...