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

[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.)

 

[Mentallic]

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

 

[Tyrion101]

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

 

[Mentallic]

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

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

[tex]x^2-\frac{1}{x}-1[/tex]

and then factor out 1/x giving us

[tex]\frac{1}{x}\left(x^3-x-1\right)[/tex]

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

[tex]\tan^2{x}=\sec^2{x}-1[/tex]

so then use this to simplify your expression.

 

[Tyrion101]

This is my problem.

 

[Tyrion101]

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

 

[Mentallic]

Please use parentheses in future.

x^2-1/x-1
is read as
[tex]x^2-\frac{1}{x}-1[/tex]

while
(x^2-1)/(x-1)
[tex]=\frac{x^2-1}{x-1}[/tex]

 

[HallsofIvy]

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