Simplify this to one trigonometric function

I'm having trouble here, how do you do this and what is the answer? (explain the steps, i've got a test i'm preparing for in a few days)

(sin^2x)(cot^2x) = ?

cosx(1+tan^2x) = ?

cosx / 1 - sinx - cosx / 1 + sinx = ?

1.)

(sin^2x)(cot^2x) = (sin^2x) (cos^2x/sin^2x) = cos^2x

2.)

cosx(1+tan^2x) = cosx( sec^2x) = cosx (1/cos^2x) = 1/cosx = secx


3.)

cosx / (1 - sinx) - cosx / (1 + sinx) = ...


(1+sinx)cosx / (1 - sinx)(1+sinx) - (1-sinx)cosx / (1 + sinx)(1-sinx)

{ we are getting a common deminator }

cosx(1+sinx)/(1-sin^2x) - (1-sinx)cosx/(1-sin^2x) =

[cosx(1+sinx) - cosx(1-sinx)] / (1-sin^2x)=

[cosx(1+sinx) - cosx(1-sinx)] / cos^2x=

[(1+sinx)-(1-sinx) ]/cosx =

2sinx/cosx =

2 tanx

Rules
tan = sin/cos
cot = cos/sin
sin^2 + cos^2 = 1

Bchbhx