Help with trig identities

How do I prove this:
(cscx/cscx - 1) + (cscx/cscx+1) = 2sec^2x.

Major test tmrw so any help is appreciated.

I am assuming that the expression is:

cscx/(cscx - 1) + cscx/(cscx + 1) = 2sec²x.

Since csc(x) = 1/sin(x):

csc(x)/(csc(x) - 1) + csc(x)/(csc(x) + 1)
= 1/sin(x)/(1/sin(x) - 1) + 1/sin(x)/(1/sin(x) + 1)

Then, for each fractional expression, multiply the numerator and denominator of sin(x).

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

So the expression becomes:

1/(1 - sin(x)) + 1/(1 + sin(x))

Next, by LCD method and expanding (1 + sin(x)(1 - sin(x)):

1/(1 - sin(x)) * (1 + sin(x))/(1 + sin(x)) + 1/(1 + sin(x)) * (1 - sin(x))/(1 - sin(x))
= (1 + sin(x))/((1 + sin(x))(1 - sin(x))) + (1 - sin(x))/((1 + sin(x))(1 - sin(x)))
= (1 + sin(x) + 1 - sin(x))/((1 + sin(x))(1 - sin(x)))
= 2/(1 - sin²(x))

Recall that the Pythagorean identity is:

sin²(x) + cos²(x) = 1

Rearrange the terms:

1 - sin²(x) = cos²(x)

So:

2/cos²(x)
= 2sec²(x) [Noting that 1/cos²(x) = sec²(x)]
= RHS

I hope this helps!

I'm assuming
(csc / [csc - 1]) + (csc/[csc + 1]) = csc ([csc + 1] + [csc - 1])/(csc^2 - 1) = 2csc^2/(csc^2 - 1) = 2 / ( 1 - sin^2) = 2/cos^2 = 2 sec^2

csc = 1/sin, sec = 1/cos