Verify: cotx = (sin2x)/(1-cos2x)

The right side is sin2x/(1-cos2x)

But Sin (a+b) = sinacosb+cosasinb.

Therefore, sin2x = 2sinxcosx...........(1)

Also we know that cos(a+b) = cosacosb-sinasinb. S0 co2x = cos^2x-sin^2x = (1-sin^2x-sin^2x) = 1-2sin^2x.........(2)

So, RHS by virtue of (1) and(2)  is,

2sinxcosx/ {1- (1-2sin^2 x)} = 2sinxcosx/(2sin^2 x) = cosx/sinx = cotx equal to LHs