# Tan^2x =sec^2x-1 also: tan= sec -1 or am I missing something?

1. Feb 17, 2015

### Tyrion101

A particular problem with factoring has both of these, one in the denominator and one in the numerator, if it were algebra it would look like: x^2-1/x-1. The trouble is I've forgotten how to simplify this. I'm on taptalk.)

2. Feb 17, 2015

### Mentallic

Can you please make more sense of this. What is your problem exactly? And no, $\tan{x}\neq \sec{x}-1$

3. Feb 17, 2015

### Tyrion101

That just looks like gibberish on taptalk. Essentially it is Sec^2x-1/Secx-1 once a bit of factoring is done.

4. Feb 17, 2015

### Mentallic

So you want to simplify this expression by factoring or any other means? Ok well, if we take a look at

$$x^2-\frac{1}{x}-1$$

and then factor out 1/x giving us

$$\frac{1}{x}\left(x^3-x-1\right)$$

The cubic has no rational factors, so that is the best we can do. We're not completely at a loss with simplifying though.

You know that

$$\tan^2{x}=\sec^2{x}-1$$

so then use this to simplify your expression.

5. Feb 17, 2015

### Tyrion101

This is my problem.

6. Feb 17, 2015

### Tyrion101

I solved my problem... I was not removing the square when factoring. Always seems to be that kind of mistake that gets me.

7. Feb 17, 2015

### Mentallic

x^2-1/x-1
$$x^2-\frac{1}{x}-1$$

while
(x^2-1)/(x-1)
$$=\frac{x^2-1}{x-1}$$

8. Feb 19, 2015

### HallsofIvy

$$\frac{x^2- 1}{x- 1}= \frac{(x- 1)(x+ 1)}{x- 1}= x+ 1$$ as long as x is not equal to 1.