Find the first 3 non - zero term of Taylor series expansion of f(x) = x/cos(x) at x = 0

can you write the expansion and explain in detail pls, thanks so much

Note that x/cos(x) is odd since:
f(-x) = -x/cos(-x) = -x/cos(x) = -f(x),

and recall that the Taylor Series expansion of an odd function at x = 0 contains only odd-degreed terms. Thus, for some A, B, and C, we can write:
x/cos(x) = Ax + Bx^3 + Cx^5 + ... .
(Note: there is no constant term since f(0) = 0.)

The Taylor series for cos(x) at x = 0 is:
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ... ,

so we have:
x/(1 - x^2/2! + x^4/4! - x^6/6! + ...) = Ax + Bx^3 + Cx^5 + ...
==> x = (1 - x^2/2! + x^4/4! - x^6/6! + ...)(Ax + Bx^3 + Cx^5 + ...).

If we expand the RHS and collect like terms, we get:
x = Ax + (-A/2! + B)x^3 + (A/4! - B/2! + C)x^5 + ... .

(i) Comparing the x coefficients: A = 1
(ii) Comparing x^3 coefficients: -A/2! + B = 0 ==> B = A/2 = 1/2
(iii) Comparing x^5 coefficients: A/4! - B/2! + C = 0 ==> C = -A/4! + B/2! = 5/24.

Therefore:
x/cos(x) = x + (1/2)x^3 + (5/24)x^5 + ... .

I hope this helps!

I believe it's x+((x^3)/2+(5x^5)/24 use the formula f(a) + f'(a)(x) + f'(a)(x^2)/2! In other words find the derivative and plug it into the formula, the general term would be f^(n)(x^n)/(n!) where n is the number of derivatives and ! is the factorial.