shows that sinx tanx maybe written as 1-cos2x / c osx?

Trigonometric identities used:
• tanθ = sinθ / cosθ
• sin²θ + cos²θ = 1  and thus  sin²θ = 1 - cos²θ

L.H.S.
= sinx tanx
= sinx (sinx / cosx)
= sin²x / cosx
= (1 - cos²x) / cosx
= R.H.S.

Hence, sinx tanx = (1 - cos²x)/cosx

sinx.tanx = sinx(sinx/cosx)

i.e. sin²x/cosx

Now, you say cos2x instead of cos²x....??

Now, cos2x = 1 - 2sin²x

so, 2sin²x = 1 - cos2x

Then, sin²x = (1 - cos2x)/2

Hence, sinx.tanx => (1 - cos2x)/2cosx    

Did you intended to say (1 - cos²x)/cosx...?

Well sin²x + cos²x = 1

so, sin²x = 1 - cos²x

i.e. sinx.tanx => (1 - cos²x)/cosx

:)>

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