Try writing cos(x) * tan(x) = sin(x)
Now (1-x^2/2!+x^4/4!...) * (a+bx+cx^2+dx^3+ex^4+fx^5...) = (x-x^3/3!+x^5/5!)
You can equate coefficients e.g. the same number of x's have to appear on each side of the equation.
So, firstly there are no constants on the right, so a=0 (because a*1 in the expansion of the LHS would give you a constant).
Secondly, bx*1 = x, therefore b=1.
Thirdly, cx^2*1 + a*(-x^2/2!) = 0, therefore c=0 (because we found a=0).
Fourth, dx^3*1 + (bx)*(-x^2/2!) = -x^3/3!. Solving this gives you d = 1/3. (We found b=1).
And so on. Do it for as many terms as you feel necessary (perhaps up to x^5).
I must admit, I don't know a neater analytic way to do it.