Question

Question #8de97

Answer 1

That is not an identity.

Explanation:

Recall that

`cot^2x+1 = csc^2x`.

So, we can write

`(1-csc^2x)/csc^2x = (1-(cot^2x+1))/csc^2x`

`= cot^2x/csc^2x`

Recall also that `cotx = cosx/sinx` and `cscx = 1/sinx`.

This allows us to continue

`= (cos^2x/sin^2x)/(1/sin^2x)`

`= cos^2x/sin^2x * sin^2x/1`

`= cos^2x`

Which is not identically `cosx`.

(`cos^2x = cosx` only when `cosx = 1` or `0`)

Answer 2

No. It is equal to `sin^2x - 1`.

Explanation:

If we have `(1 - x)/x`, we can write it as `1/x - x/x`.

The same way, `(1 - csc^2 x)/csc^2` can be written as `1/csc^2x - (csc^2x)/(csc^2x)`.

This is equal to `sin^2x - 1`.