Question #fdae9

1 Answer
May 13, 2017

I tried this:

Explanation:

We can use the fact that:

tanx=sinx/cosxtanx=sinxcosx

and

sin^2x+cos^2x=1sin2x+cos2x=1

and rearrange to write:

cos(x)[1/(1-sinx)-1/(1+sinx)]=2sinx/cosxcos(x)[11sinx11+sinx]=2sinxcosx

cos(x)[(cancel(1)+sinxcancel(-1)+sinx)/((1-sinx)(1+sinx))]=2sinx/cosx

cos(x)[(2sinx)/(1-sin^2x)]=2sinx/cosx

and cancel:

cancel(cos(x))[(2sinx)/(cos^cancel(2)x)]=2sinx/(cosx)

therefore2sinx/cosx=2sinx/cosx