1-tan(x)/1+tan(x) = 1-sin2x/cos2x

Show your attempt please.

The way you have stated this problem, I do not think it is possible to prove it
as a valid trigonometric identity. Did you mean the following:

{1-tan(x)} {1+tan(x)} = 1 - {(sin(x))^2 / (cos(x))^2}

The above equation is read, "The quantity one minus tangent x, times the
quantity one plus tangent x, equals one minus the ratio sine x, divided by
cosine x, the ratio raised to the 2nd power." Is that what you meant?

If so, the solution is as follows:

Recall that sin(x) / cos(x) = tan(x), therefore, the right side of the above
equation becomes:

1 - {(sin(x))^2 / (cos(x))^2} = 1 - {tan(x)}^2

The right side of the above equation is "the difference of two squares."
Recall that the difference of two squares can be factored as follows:

(a^2 - b^2) = (a-b)(a+b)

In our case, a = 1, and b = tan(x); therefore,

1 - {tan(x)}^2 = 1^2 - {tan(x)}^2 = (1 - tan(x)) (1 + tan(x)) Q.E.D.

"Q.E.D." means "Quod Erat Demonstrandum," Latin for "which was to be
demonstrated." The initials are used often in math textbooks.