Question

How do you prove `(sinx + cosxcotx)/cotx = secx`?

Answer 1

We need to get the left side equal to the right side:

So we will look at `=(sinx + cosxcotx)/cotx`

we can use the identity of `color(red)(cotx = cosx/sinx)` here

so we will have `=(sinx + cosx(cosx/sinx))/(cosx/sinx)`

if we now take out a `color(red)(1/sinx)` from the top, we will have:

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

there we can use the identity that `color(red)(sin^2x + cos^2x = 1)`

thus we are left with `=(1/sinx)/(cosx/sinx)`

which we can write as `=(1/sinx)(sinx/cosx)`

the `color(red)(sinx)`'s will cancel each other, and we will be left with:

`= (1/cosx)`

which is our identity to `color(red)((1/cosx) = secx)`

thus we get `=secx`

and we have that the right hand side `=` to left hand side