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

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The discussion focuses on simplifying the expression involving trigonometric identities, specifically the relationship between tangent and secant functions, where it is established that tan²x = sec²x - 1. A common mistake identified is the failure to properly factor expressions, particularly when dealing with (x² - 1)/(x - 1). The correct simplification leads to x + 1, provided x ≠ 1. The importance of using parentheses for clarity in mathematical expressions is emphasized.

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Tyrion101
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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.)
 
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Can you please make more sense of this. What is your problem exactly? And no, \tan{x}\neq \sec{x}-1
 
That just looks like gibberish on taptalk. Essentially it is Sec^2x-1/Secx-1 once a bit of factoring is done.
 
Tyrion101 said:
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

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.
 
ImageUploadedByTapatalk1424221100.710027.jpg

This is my problem.
 
I solved my problem... I was not removing the square when factoring. Always seems to be that kind of mistake that gets me.
 
Please use parentheses in future.

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

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