Question

Prove that `(cosxcotx)/(1 - sinx) - 1 = cscx`?

Answer 1

We start with:

`color(green)((cosxcotx)/(1-sinx) - 1 stackrel(?)(=) cscx).`

First, recall `cotx = cosx/sinx` (since `cotx = 1/tanx` and `tanx = sinx/cosx`). That gives you:

`(cosx*cosx/sinx)/(1-sinx) - 1 = cscx`

`(cos^2x/sinx)/(1-sinx) - 1 = cscx`

On the fraction you can move the middle `sinx` into the denominator.

`cos^2x/(sinx(1-sinx)) - 1 = cscx`

Now, when you have a `1`, that gives you a lot of freedom. You can choose to have it be equal to any ratio you want, as long as it cancels out to be `1`. So choose `1 = (sinx(1-sinx))/(sinx(1-sinx))` to get:

`cos^2x/(sinx(1-sinx)) - (sinx(1-sinx))/(sinx(1-sinx)) = cscx`

Distribute the numerator and combine into one fraction:

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

Then recall `sin^2x + cos^2x = 1` to cancel out the `1-sinx`.

`cancel(1 - sinx)/(sinxcancel((1-sinx))) = cscx`

`1/sinx = cscx`

`color(blue)(cscx = cscx)`

Answer 2

It is a bit long but...

Explanation:

You can try changing all in `sin` and `cos` as:
`cot(x)=cos(x)/sin(x)`
`csc(x)=1/sin(x)`
Your identity becomes:

`(cos(x)*cos(x)/sin(x))/(1-sin(x))-1=1/sin(x)`
we use `(1-sin(x))(sin(x))` as common denominator and write rearranging:
`(cos^2(x)-sin(x)(1-sin(x)))/cancel((1-sin(x))(sin(x)))=(1-sin(x))/cancel((1-sin(x))(sin(x)))`

and:
`cos^2(x)-sin(x)(1-sin(x))=1-sin(x)`
`cos^2(x)-sin(x)+sin^2(x)=1-sin(x)`
but:
`cos^2(x)+sin^2(x)=1`
so we get:
`1-sin(x)=1-sin(x)` which is true.