f(b)=(b-b^2)/(1+b^2), f(tan x)

$$\\f(tan(x))=\frac{tanx-tan^2x}{1+tan^2x}\\\\
f(tan(x))=\frac{tanx(1-tanx)}{sec^2x}\\\\
f(tan(x))=cos^2x\;tanx(1-tanx)\\\\
f(tan(x))=\frac{cos^2xsinx}{cosx}(1-tanx)\\\\
f(tan(x))=cosxsinx(1-tanx)\\\\
f(tan(x))=cosxsinx-sin^2x\\\\
f(tan(x))=sin(x)[cos(x)-sin(x)]\\\\
or\\
f(tan(x))=\frac{sin2x-2sin^2x}{2}\\\\$$

Edited:   Thanks fiora

f(b)=(b-b^2)/(1+b^2),

f(tan x)=[tan(x)-tan^2(x)]/[cos^2(x)/cos^2(x)+sin^2(x)/cos^2(x)]

           =[tan(x)-tan^2(x)]/[1/cos^2(x)]

           =[tan(x)-tan^2(x)]*cos^2(x)

           =[sin(x)/cos(x)-sin^2(x)/cos^2(x)]*cos^2(x)

           =sin(x)cos(x)-sin^2(x) or =1/2*sin2x+1/2*cos2x-1/2

Hi,Melody.

I think there is something wrong in anwser.

In your second step,f(tan(x))=tanx*(1-tanx)/sec^2(x)

                                         =tanx*cos^2(x)*(1-tanx)

                                         =cosx*sinx*(1-tanx)

You were right fiora - I had made a mistake.  Does it look right now?   Thank you  

Yes.I believe it's correct.

Thaks fiora :)