How to prove 1/(secx+1)=cotx(cscx-cotx)

1/(secx+1)=cotx(cscx-cotx)

and

4/(csc^2x+sec^2x=sin^2 2x)

1)
1/(secx+1) = cotx(cscx-cotx)

R.H.S. = cotx(cscx-cotx) = (cosx/sinx)[1/sinx -cosx/sinx]
R.H.S. = cosx(1-cosx)/sin^2x = cosx(1-cosx) / (1-cos^2x)
R.H.S. = cosx / (1+cosx)

R.H.S. = 1 / (secx + 1) = L.H.S >===========< Q . E . D.

2)
4/(csc^2x+sec^2x) = sin^2 2x

L.H.S. = 4/(csc^2x+sec^2x) = 4/[1/sin^2x + 1/cos^2x]
L.H.S. = 4sin^2x.cos^2x / (cos^2x + sin^2x)

L.H.S. = (2sinx.cosx)^2

L.H.S. = sin^2 2x = R.H.S. >=============< Q . E .D.

1/(sec x + 1) = cot x(csc x - cot x)

Simplifying RHS:

(cos x/sin x)*(1/sin x - cos x/sin x)

(cos x)(1 - cos x)/(1 - cos² x) = cos(x)/(1 + cos x) = 1/sec x/[1 + 1/sec x] = 1/(sec x + 1)