Show that the derivative of `(sinx)/(1 - cosx)` is `-cscx`?

`sinx/(1-cosx)`

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

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

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

= `(1+cosx)/sinx`

= `cscx+cotx`

As such `d/(dx)(sinx/(1-cosx))`

= `d/(dx)(cscx+cotx)`

= `d/(dx)cscx+d/(dx)cotx`

= `-cscxcotx-csc^2x`

We can't, because it's false...

Let's use the product rule.

`d/(dx)[(sinx)/(1-cosx)]`

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

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

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

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

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

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

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

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

`= -1/(1-cosx)`

But we know that `-1/sinx = -cscx`. Therefore, we claim that

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

Therefore:

`sinx/(1-cosx) = 1`

`sinx = 1 - cosx`

`sinx + cosx ne 1`

Clearly, we have found that this identity is false. More definitively, we can plot this equation to get:

`sinx + cosx`:

graph{sinx + cosx [-10, 10, -5, 5]}

As this equation is not a straight horizontal line, it cannot be equal to `1` for all `x` in `RR`.