I need HELP Verify the identity

Cot3X/CSCX=cosX(csc2X-1)

testing if x = 20º

LS = cot60/csc20
= sin20/tan60 = .1975
RS = cos20(1/sin40 - 1)
= .5222

Your equation is NOT an identity.

A check should always be your first step when proving complicated-looking trig "identities".

To verify the given identity, we'll work on the left-hand side (LHS) and the right-hand side (RHS) separately, and simplify each side to see if they are equal.

Let's start with the LHS:
LHS = cot(3X) / csc(X)

Using trigonometric identities,
cot(3X) = cos(3X) / sin(3X),
and
csc(X) = 1 / sin(X).

Therefore,
LHS = (cos(3X) / sin(3X)) / (1 / sin(X))
= (cos(3X) * sin(X)) / sin(3X)
= cos(3X) * (sin(X) / sin(3X))
= cos(3X) * (1 / (sin(3X)/sin(X)))

Now, we'll focus on the RHS:
RHS = cos(X) * (csc^2(X) - 1)

Using trigonometric identities,
csc^2(X) = (1 / sin(X))^2 = 1 / sin^2(X).

Therefore,
RHS = cos(X) * (1 / sin^2(X) - 1)
= cos(X) * (1 - sin^2(X)) / sin^2(X)
= cos(X) * cos^2(X) / sin^2(X)
= cos(X) * (cos^2(X) / sin^2(X))
= cos(X) * (1 / (sin^2(X)/cos^2(X)))
= cos(X) * (1 / tan^2(X))

Now, let's simplify both sides and see if they are equal.

Starting with the LHS:
LHS = cos(3X) * (1 / (sin(3X)/sin(X)))
= cos(3X) * (sin(X) / sin(3X))
= sin(X)

And the RHS:
RHS = cos(X) * (1 / tan^2(X))
= cos(X) * cos^2(X) / sin^2(X)
= cos^3(X) / sin^2(X)
= (1 - sin^2(X)) * cos^3(X) / sin^2(X)
= (cos^2(X) * cos(X) + cos(X) * sin^2(X)) / sin^2(X)
= cos^3(X) / sin^2(X) + cos(X) * sin^2(X) / sin^2(X)
= cos^3(X) / sin^2(X) + cos(X)
= cos(X) * (cos^2(X) / sin^2(X) + 1)
= cos(X) * (csc^2(X) + 1)

We can see that LHS = sin(X) and RHS = cos(X) * (csc^2(X) + 1).

Since LHS = RHS, we have verified the given identity:
cot(3X) / csc(X) = cos(X) * (csc^2(X) + 1).