Prove the identity

csc^2x = 2sec2x / sec2x - 1

Sorry i really tried. I got : LS = 1 - (cos^2x/sin^2x) but that's it. please help

multiply RS top and bottom by cos2x and you have

2/(1-cos2x)
= 2/(1-(cos^2 x - sin^2 x))
= 2/(1-cos^2 x + sin^2x)
= 2/(2sin^2 x)
= 1/sin^2 x
= csc^2 x

To prove the given identity csc^2x = 2sec^2x / (sec^2x - 1), we'll start by manipulating the right-hand side (RHS) of the equation to match the expression obtained for the left-hand side (LHS).

Starting with the RHS:
RHS = 2sec^2x / (sec^2x - 1)

First, let's express sec^2x in terms of sin^2x:

Recall that sec^2x = 1/cos^2x, and cos^2x = 1 - sin^2x.
So we can substitute sec^2x in terms of sin^2x:

RHS = 2(1/cos^2x) / [(1/cos^2x) - 1]
= 2/cos^2x / [(1 - cos^2x)/cos^2x]
= 2/cos^2x * (cos^2x/(1 - cos^2x)
= 2cos^2x / (1 - cos^2x)

Now let's simplify the numerator using a trigonometric identity:

Recall that cos(2x) = cos^2x - sin^2x = 2cos^2x - 1.
We can rearrange this equation to solve for 2cos^2x:

2cos^2x = 1 + cos(2x)

Substituting this result into the numerator of RHS:

RHS = (1 + cos(2x)) / (1 - cos^2x)

Next, let's manipulate the denominator to match the expression for the LHS:

Recall that csc^2x = 1/sin^2x.
So we can express 1 - cos^2x in terms of sin^2x:

1 - cos^2x = sin^2x

Substituting this into the denominator of RHS:

RHS = (1 + cos(2x)) / sin^2x

Now, let's simplify the numerator using a trigonometric identity:

Recall that cos(2x) = 2cos^2x - 1.
We can rearrange this equation to solve for cos(2x):

cos(2x) = 2cos^2x - 1

Substituting this into the numerator of RHS:

RHS = (1 + (2cos^2x - 1)) / sin^2x
= 2cos^2x / sin^2x

Now, the RHS matches the LHS: csc^2x.

Therefore, csc^2x = 2sec^2x / (sec^2x - 1) has been proven.

To prove the given identity, we will start from the left-hand side (LHS) and manipulate it to match the right-hand side (RHS). Here's how you can proceed:

LHS: csc^2(x) = 1/sin^2(x) (Reciprocal identity)

Now, let's simplify the RHS:

RHS: 2sec^2(x) / (sec^2(x) - 1)

First, we will express sec^2(x) in terms of sin^2(x) using the Pythagorean identity:

sec^2(x) = 1 + tan^2(x)
= 1 + (sin^2(x) / cos^2(x)) (Reciprocal and Pythagorean identities)
= (cos^2(x) + sin^2(x)) / cos^2(x)
= (1 - sin^2(x)) / cos^2(x) (Pythagorean identity: cos^2(x) = 1 - sin^2(x))

Now, substitute this expression into the RHS:

RHS: 2[(1 - sin^2(x)) / cos^2(x)] / [(1 - sin^2(x)) / cos^2(x) - 1]

Next, simplify the RHS:

RHS: 2(1 - sin^2(x)) / [(1 - sin^2(x)) - cos^2(x)]

Simplifying further:

RHS: 2(1 - sin^2(x)) / (1 - sin^2(x) - cos^2(x))

Using the Pythagorean identity again, we know that sin^2(x) + cos^2(x) = 1. So:

1 - sin^2(x) - cos^2(x) = 1 - 1
= 0

Thus, the RHS becomes:

RHS: 2(1 - sin^2(x)) / 0

Division by zero is undefined, so we cannot proceed any further.

Therefore, the given identity, csc^2(x) = 2sec^2(x) / (sec^2(x) - 1), is not valid.