# Need help to prove this Identity!

1. Jul 14, 2010

### klmathlover

Help! I spent 3 hours attempting this question. Prove the following identity :

(tan x + sec x -1) / (tan x - sec x + 1) = tan x + sec x

I've simplified Left Hand Side into cos and sine. Which ended up like this
(sine x - cos x + 1) / (sine x + cos x -1)

Then I'm stuck.

Any help is very much appreciated!

2. Jul 15, 2010

### Staff: Mentor

So far, so good.
Now multiply by 1 in the form of (sin x + cos x + 1)/(sin x + cos x + 1). Remember the identity sin2x + cos2x = 1.

Last edited: Jul 15, 2010
3. Jul 15, 2010

### klmathlover

I've done the multiplication and now I ended up with (sin^2 x - cos^2 x - 1) / (2 + 2sinxcosx -2sinx)

I tried to further simplify into sin^2 x - 1 / 1 + sinxcosx - sin x, but then I'm stuck here.. wondering if this is the correct method.

4. Jul 15, 2010

### Staff: Mentor

Sorry, I steered you wrong. My scratch work was correct but I copied a sign incorrectly in my post. The left side of the identity is equal to (sin x + 1 - cos x)/(sin x - 1 + cos x), which equals (sin x - cos x + 1)/(sin x + cos x - 1). Multiply this expression by 1 in the form of (sin x + cos x + 1)/(sin x + cos x + 1).

5. Jul 15, 2010

### klmathlover

Thanks - I've got it. :)

May I ask you - how do you know what to do with this question in the first place? I mean, how do you know to mulitply by one (which I understand and make sense) but how to do you what to multiply with?

I mean it could be sinx + cosx +1 or sinx - cos x + 1 or any other combination.

Do you have a specific method or have to try out one by one?

6. Jul 16, 2010

### Staff: Mentor

Sometimes you have to try different things out, but it boils down to multiplying a + b by a - b to get a2 - b2, or multiplying a - b by a + b to get the same thing, to see if that gets you anywhere. With those squared terms I was hoping to get sin2(x) + cos2(x), which I can replace by 1.

In your problem, there was (sin x - cos x + 1)/(sin x + cos x - 1), so I looked at the denominator as a difference, (sin x + cos x) - 1, and thought multiplying by a sum, (sin x + cos x) + 1 might be useful. If that hadn't borne fruit, I would have tried a similar thing with the numerator. Of course, I can't just multiply the denominator or numerator alone, but I can always multiply by 1.

Partly it's a matter of experience, and seeing the kinds of things that worked in other problems.

7. Jul 16, 2010

### klmathlover

Thanks a lot. I can understand now. :)

8. Jul 16, 2010

### Hurkyl

Staff Emeritus
It's generally a lot easier to simplify the equation as a whole, rather than try and work with the two sides independently.