Question

Question #fdae9

Answer 1

I tried this:

Explanation:

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`