Proving Equations: cscθ - cotθ = (cscθ + cotθ)^-1

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Discussion Overview

The discussion revolves around proving the equivalence of the equations (cscθ - cotθ) and (cscθ + cotθ)^{-1}. Participants explore different methods for simplifying and proving this relationship, focusing on algebraic manipulation and trigonometric identities.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests multiplying the two sides together to show that the result is 1, indicating that this is sufficient for the proof.
  • Another participant expresses difficulty in simplifying each side independently and requests a step-by-step approach to the proof.
  • A different approach is proposed where the right side is rewritten as 1/(cscθ + cotθ) and further simplified to show it equals cscθ - cotθ.
  • One participant corrects another by stating that equations can be equivalent but not equal, emphasizing the distinction in terminology.
  • Another participant suggests converting to sine and cosine for simplification, providing a detailed algebraic manipulation to demonstrate the equivalence.
  • There is a back-and-forth regarding whether participants misread the question, with assertions that their interpretations and methods are correct.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to prove the equivalence, with some advocating for multiplication while others prefer direct simplification. There is no consensus on a single method, and the discussion remains unresolved regarding the most effective proof strategy.

Contextual Notes

Some participants highlight potential misunderstandings in the interpretation of the question, and there are varying levels of detail in the algebraic steps provided. The discussion reflects different mathematical perspectives and approaches to the problem.

MathewsMD
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I am trying to prove that the equations below equal each other.

## (cscθ - cotθ) = (cscθ + cotθ)^{-1} ##

Besides letting the two sides equal each other, I have not been able to simplify each side independently to find the equivalent expression. I've had to remove terms from either side by multiplying each side simultaneously, but if anyone could show me how to do this proof one equation at a time, then that would be great!
 
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Just multiply them together and note that the result is 1. That is sufficient.
 
MathewsMD said:
I am trying to prove that the equations below equal each other.

## (cscθ - cotθ) = (cscθ + cotθ)^{-1} ##

Besides letting the two sides equal each other, I have not been able to simplify each side independently to find the equivalent expression. I've had to remove terms from either side by multiplying each side simultaneously, but if anyone could show me how to do this proof one equation at a time, then that would be great!
Write the right side as 1/(cscθ + cotθ) which can be rewritten as
$$ \frac{1}{\frac{1}{sinθ} + \frac{cosθ}{sinθ}}$$
A bit more work shows that the last expression above is equal to cscθ - cotθ.

A quibble with your first sentence, above. One equation does not "equal" another. Equations can be equivalent (meaning they have exactly the same solution set), but never equal to one another.
 
jgens said:
Just multiply them together and note that the result is 1. That is sufficient.
Did you misread the question? Multiplying them together gives
[tex]\frac{csc(\theta)- cot(\theta)}{csc(\theta)+ cot(\theta)}[/tex]
which can be shown to be equal to 1 but is not obviously so. Did you miss the sign change?

With problems of this kind I almost automatically change to "sine" and "cosine".
[tex]csc(\theta)- cot(\theta)= \frac{1}{sin(\theta}- \frac{cos(\theta)}{sin(\theta)}[/tex]
[tex]= \frac{1- cos(\theta)}{sin(\theta)}[/tex]
Multiply both numerator and denominator by [itex]1+ cos(\theta)[/itex]:
[tex]= \frac{1- cos^2(\theta)}{sin(\theta)(1+ cos(\theta))}[/tex]
 
HallsofIvy said:
Did you misread the question?

Nope. My suggestion: (csc(x)-cot(x))(csc(x)+cot(x)) = csc2(x)-cot2(x) = csc2(x)(1-cos2(x)) = csc2(x)sin2(x) = 1. It seemed like the most straight-forward solution to me.

Did you miss the sign change?

Again no.
 

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