Verify:

cotx = (sin2x)/(1-cos2x)

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tanx = (sin2x)/(1-cos2x)

= (sin2x)/2sin^2x

= 2sinxcosx/(2sin^2x)

= cosx/sinx

= cotx therefore not an identity

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

cotx=(sin2x)/(1-cos2x)

(sin2x)/(1-cos2x)=(2sinxcosx)/(1-cos2x) : Double Angle Formula sin2x=2sinxcosx

(2sinxcosx)/(2sin^2x) : Power Reducing Formula sin^2x=(1-cos2x)/(2)

cosx/sinx : Divide by 2sinx/2sinx

cosx/sinx=cotx : Quotient Identity

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