Parametric Equations help

The question is as follows:

The parametric equations of a curve, C are:
2x = cot(u) - cosec(u) and 4y = cosec(u) - 2cot(u)

(i) Express cot(u) and cosec(u) in terms of x and y

(ii) Find an equation connecting x and y

(i)
2x = cot(u) - cosec(u) ...(1)
4y = cosec(u) - 2cot(u) ...(2)

Multiply equation (1) by 2 and then add it to the equation (2)

4x = 2cot(u) - 2cosec(u)
4y = -2cot(u) + cosec(u)
-----------------------------------
4x + 4y = -cosec(u)

or cosec(u) = -4(x+y)

plug this into (1) or (2) to find cot (u)

2x = cot(u) - cosec(u)
2x = cot(u) +4x + 4y
cot(u) = -2x - 4y = -2(x+2y)

(ii) To connect y to x, use the identity

1+cot²(u) = cosec²(u)

Using the expressions found for cot(u) and cosec(u), you get

1+ [-2(x+2y)] ²=[-4(x+y)]²
or
1+ 4(x² + 4y² + 4xy) = 16(x²+y²+2xy)
or
12x² + 16xy = 1

If we add 2x+4y:
2x+4y=cot(u) - csc(u)+csc(u)-2cot(u)
so 2x+4y=-cot(u) so cot(u)=-2x-4y

if we add 4x+4y we get:
4x+4y=2(cot(u)-csc(u))+csc(u)-2cot(u) = 2cot(u)-2csc(u)+csc(u)=2cot(u)
so 4x+4y=-csc(u), so csc(u)=-4x-4y

to conne t them: cot^2+1=csc^2
so 4x^2+16y^2+16xy+1=16x^2+16y^2+32xy
so 1=12x^2+16xy

(i) Add the two equations and get 2x+4y = -cot(u)
Substitute in the 1st and get 2x = -2x-4y -cosec(u) => cosec(u) = -4(x+y)

(ii) You do it