Question #fdae9

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Answer 1 · 2017-05-13T23:41:39

I tried this:

We can use the fact that:

#tanx=sinx/cosx#

and

#sin^2x+cos^2x=1#

and rearrange to write:

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

#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#