How do you show that `(cscx - cotx)^2 = (1 - cosx)/(1 + cosx)`?

Convert to sine and cosine using `cscx = 1/sinx` and `cotx= cosx/sinx`.

`(1/sinx - cosx/sinx)^2 = (1 - cosx)/(1 + cosx)`

`((1 - cosx)/sinx)^2 = (1 - cosx)/(1 + cosx)`

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

Now use `sin^2x+ cos^2x = 1`.

`(1 - cosx)^2/(1 - cos^2x) = (1 - cosx)/(1 + cosx)`

Note the difference of squares in `1 - cos^2x`, such that `1 - cos^2x = (1 + cosx)(1- cosx)`.

`(1- cosx)^2/((1 + cosx)(1 - cosx)) = (1 - cosx)/(1 + cosx)`

`(1 - cosx)/(1 + cosx) = (1 - cosx)/(1 + cosx)`

This identity has been proven to be true.

Hopefully this helps!

Whenever you see a trig identity proof that has something like `(1 +- "trig " x)` in a denominator (where "trig" is one of the 6 trig functions), a good first thing to try is multiplying it by its conjugate.

In our case, `(1+cosx)` is in a denominator, so we choose to multiply it by its conjugate, `(1-cosx)`. (To do this, of course, we need to multiply the numerator by it as well, so we're essentially multiplying the fraction by a fancy version of `1`, which does not change its value.)

`(1-cosx)/(1+cosx)=(1-cosx)/(1+cosx) color(red)(xx (1-cosx)/(1-cosx))`

`color(white)((1-cosx)/(1+cosx))=(1-cosx)^2/(1-cos^2x)"             "`(distribution)

`color(white)((1-cosx)/(1+cosx))=(1-cosx)^2/(sin^2x)"             "`(Pythagorean identity)

`color(white)((1-cosx)/(1+cosx))=((1-cosx)/sinx)^2"        "["thing"^2/"stuff"^2=("thing"/"stuff")^2]`

`color(white)((1-cosx)/(1+cosx))=(1/sinx-cosx/sinx)^2"     "`(splitting the fraction)

`color(white)((1-cosx)/(1+cosx))=(cscx-cotx)^2"           "`(reciprocal identities)

Note:

You can also convert everything to `sin`'s and `cos`'s. This is sort of a "catch-all", like how the quadratic formula can be used to solve any quadratic equation (of one variable). It will always work, but there might be a faster way.