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

  • Thread starter MathewsMD
  • Start date
Multiplying both sides by (csc(x)+cot(x)) gives:(csc(x)-cot(x))(csc(x)+cot(x)) = (csc(x)+cot(x))(csc(x)+cot(x))(csc(x)-cot(x))= (csc2(x)+cot2(x))(csc(x)-cot(x)) = 1(csc(x)-cot(x)) = csc(x)-cot(x).Therefore, (csc(x)-cot(x)) = (csc(x)+cot(x))^(-1).In summary, the equations (cscθ - cotθ) and (cscθ + cotθ)^{-1} are equivalent and can be proven by multiplying them together and simplifying to get
  • #1
MathewsMD
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7
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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  • #2
Just multiply them together and note that the result is 1. That is sufficient.
 
  • #3
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.
 
  • #4
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]
 
  • #5
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.
 

1. What is the purpose of proving equations?

Proving equations is an important step in mathematics as it allows us to verify the validity of a given equation and ensure that it holds true for all values of the variables involved.

2. How do we prove an equation like cscθ - cotθ = (cscθ + cotθ)^-1?

To prove an equation, we need to show that both sides of the equation are equal. In this case, we can use algebraic manipulation and trigonometric identities to simplify the left side of the equation and show that it is equal to the right side.

3. Can we prove this equation using a proof by contradiction?

Yes, we can prove this equation using a proof by contradiction by assuming that the equation is false and showing that it leads to a contradiction. This would prove that the equation is, in fact, true.

4. What are some common mistakes to avoid when proving equations?

Some common mistakes to avoid when proving equations include making incorrect algebraic manipulations, using incorrect trigonometric identities, and not considering all possible values of the variables involved.

5. Are there any tips for effectively proving equations?

Yes, some tips for effectively proving equations include starting with the side that is more complex, clearly stating the steps and reasoning used, and checking the final result to ensure that it is equivalent to the original equation.

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