Question

How do you prove `sec(x) + 1 + ((1-tan^2(x)) / (sec(x)-1)) = cos(x)/(1-cos(x))`?

Answer 1

Do some conjugate multiplication, make use of trig identities, and simplify. See below.

Explanation:

Recall the Pythagorean Identity `sin^2x+cos^2x=1`. Divide both sides by `cos^2x`:
`(sin^2x+cos^2x)/cos^2x=1/cos^2x`
`->tan^2x+1=sec^2x`
We will be making use of this important identity.

Let's focus on this expression:
`secx+1`

Note that this is equivalent to `(secx+1)/1`. Multiply the top and bottom by `secx-1` (this technique is known as conjugate multiplication):
`(secx+1)/1*(secx-1)/(secx-1)`
`->((secx+1)(secx-1))/(secx-1)`
`->(sec^2x-1)/(secx-1)`

From `tan^2x+1=sec^2x`, we see that `tan^2x=sec^2x-1`. Therefore, we can replace the numerator with `tan^2x`:
`(tan^2x)/(secx-1)`

Our problem now reads:
`(tan^2x)/(secx-1)+(1-tan^2x) / (secx-1) = cosx/(1-cosx)`

We have a common denominator, so we can add the fractions on the left hand side:
`(tan^2x)/(secx-1)+(1-tan^2x) / (secx-1) = cosx/(1-cosx)`
`->(tan^2x+1-tan^2x)/(secx-1)=cosx/(1-cosx)`

The tangents cancel:
`(cancel(tan^2x)+1-cancel(tan^2x))/(secx-1)=cosx/(1-cosx)`

Leaving us with:
`1/(secx-1)=cosx/(1-cosx)`

Since `secx=1/cosx`, we can rewrite this as:
`1/(1/cosx-1)=cosx/(1-cosx)`

Adding fractions in the denominator, we see:
`1/(1/cosx-1)=cosx/(1-cosx)`
`->1/(1/cosx-(cosx)/(cosx))=cosx/(1-cosx)`

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

Using the property `1/(a/b)=b/a`, we have:
`cosx/(1-cosx)=cosx/(1-cosx)`

And that completes the proof.

Answer 2

`LHS=(secx+1)+(1-tan^2x)/(secx-1)`

`=((secx+1)(secx-1)+1-tan^2x)/(secx-1)`

`=(sec^2x-1+1-tan^2x)/(secx-1)`

`=cosx/cosx*((sec^2x-tan^2x))/((secx-1))`

`color(red)("putting",sec^2x-tan^2x=1)`

`=cosx/(cosxsecx-cosx)`

`color(red)("putting",cosxsecx=1)`

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