How do you prove `(secx)/(sinx)-(sinx)/(cosx)=cotx`?

`LHS=(secx)/(sinx)-(sinx)/(cosx)`

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

`=(1-sin^2x)/(sinxcosx)`

`=cos^2x/(sinxcosx)`

`=cosx/sinx`

`=cotx=RHS`

`secx/sinx - sinx/cosx = cotx`

I'll be simplifying down the LHS of the equation to get to the RHS of the equation. The `color(blue)("blue color")` refers to what is being changed.

First, we know that `color(blue)(sinx/cosx)` is the same as `color(blue)(tanx)`:
`secx/sinx - color(blue)(tanx)`

Now, multiply `color(blue)(sinx/sinx)` to `color(blue)(tanx)` so that both expressions have the same denominator:
`secx/sinx - (tanxcolor(blue)(sinx))/color(blue)(sinx)`

Combine both expressions to one denominator:
`(secx - tanxsinx)/sinx`

We know that `color(blue)(secx = 1/cosx)` and `color(blue)(tanx = sinx/cosx)`:
`(color(blue)(1/cosx) - color(blue)(sinx/cosx)*sinx)/sinx`

Multiply `color(blue)sinx` with `color(blue)sinx`:
`(1/cosx - color(blue)(sin^2x)/cosx)/sinx`

Combine numerator and denominator (from the top)
`(color(blue)(1-sin^2x)/cosx)/sinx`

From the Pythagorean Identities, we know that `color(blue)(1-sin^2x = cos^2x)`:
`(color(blue)(cos^2x/cosx))/sinx`

Divide `color(blue)(cos^2x/cosx)`:
`(color(blue)(cosx/1))/sinx`

Simplify:
`color(blue)(cosx)/sinx`

We know that `color(blue)(cosx/sinx = cotx)`:
`color(blue)cotx`

Now, the left hand side is equivalent to the right hand side, so we have proven that `secx/sinx - sinx/cosx = cotx`.

Hope this helps!