Trig Identities

  • #1

Homework Statement


1/1+cosx = csc^2x - cscxcotx

Homework Equations




The Attempt at a Solution


R.S= csc^2x - cscxcotx
= (1/sin^2x) - (1/sinx * cosx/sinx)
= (1/sin^2x) - (cosx/sinx^2x)
= 1-cosx / sin^2x
= 1 - cosx / 1 - cos^2x
= 1 / 1 - cos^2x

It does not match the left side and I am unsure of what I did incorrectly.
 

Answers and Replies

  • #2
963
213
= 1 - cosx / 1 - cos^2x
= 1 / 1 - cos^2x
perhaps a simple typo error

see for yourself /check
 
  • #3
perhaps a simple typo error

see for yourself /check
Where in the left side? Or in my actual work?
 
  • #4
963
213
Where in the left side? Or in my actual work?
you have on the left side
1/(1+cosx)
and on right side you got ( 1-cosx) / (1-cos^2x);
the denominator can be written as (1+cosx).(1- cosx) using identity a^2- b^2 = (a+b).(a-b)
so the numerator gets cancelled and you get 1/(1+cosx)
am i right?
 
  • #5
SteamKing
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Homework Statement


1/1+cosx = csc^2x - cscxcotx

Homework Equations




The Attempt at a Solution


R.S= csc^2x - cscxcotx
= (1/sin^2x) - (1/sinx * cosx/sinx)
= (1/sin^2x) - (cosx/sinx^2x)
= 1-cosx / sin^2x
= 1 - cosx / 1 - cos^2x
= 1 / 1 - cos^2x

It does not match the left side and I am unsure of what I did incorrectly.
This step here is the problem:

How did you go from

##=\frac{1-cos(x)}{1-cos^2(x)}## to

##=\frac{1}{1-cos^2(x)}## ?
 
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Likes drvrm
  • #6
This step here is the problem:

How did you go from

##=\frac{1-cos(x)}{1-cos^2(x)}## to

##=\frac{1}{1-cos^2(x)}## ?
I got rid of one of the cos from numerator then got rid of one from denominator.
 
  • #7
haruspex
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I got rid of one of the cos from numerator then got rid of one from denominator.
You certainly did not do that since the denominator did not change. But even if you had it would have been invalid.
In a fraction, you cannot add something to the numerator and balance that by adding the same thing to the denominator. All you can do is multiply (or divide) the numerator as a whole by something and do exactly the same to the denominator. This is what drvrm did in post #4. Even then, you do need to be a bit careful. If the thing you multiply or divide by might be zero under some circumstances then you should exclude that by writing something like "when ..... is not zero then...."
 
  • #8
You certainly did not do that since the denominator did not change. But even if you had it would have been invalid.
In a fraction, you cannot add something to the numerator and balance that by adding the same thing to the denominator. All you can do is multiply (or divide) the numerator as a whole by something and do exactly the same to the denominator. This is what drvrm did in post #4. Even then, you do need to be a bit careful. If the thing you multiply or divide by might be zero under some circumstances then you should exclude that by writing something like "when ..... is not zero then...."
Okay thank you. When doing trig identities you are able to work on both sides correct?
 
  • #9
haruspex
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Okay thank you. When doing trig identities you are able to work on both sides correct?
Not sure what you mean by that. If you mean starting with the thing to prove in the form f(x)=g(x) then applying the same operation to each side of that equation, and repeating the process until you get something that is clearly true, no. That is not a valid way to prove the original statement. The implication is in the wrong direction. It is saying "if the thing to be proved is true then ..." E.g. if you simply multiply both sides by zero you will get a true statement, but you have proved nothing.
However, that process can be used to help you find a proof. Having reached a statement that is clearly true, you can then see if all the steps you took are reversible, and hence obtain the thing to be proved.
 

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