SOLUTION: rewrite the expression in terms of the given function (1/(1+cosx))+(cosx)/(1-cosx);csc I assume this problem is asking to rewrite the equation in terms of cosecant. I do not

Algebra ->  Trigonometry-basics -> SOLUTION: rewrite the expression in terms of the given function (1/(1+cosx))+(cosx)/(1-cosx);csc I assume this problem is asking to rewrite the equation in terms of cosecant. I do not       Log On


   

Question 635264: rewrite the expression in terms of the given function
(1/(1+cosx))+(cosx)/(1-cosx);csc
I assume this problem is asking to rewrite the equation in terms of cosecant. I do not know the identities to do that though
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
1%2F%281%2Bcos%28x%29%29%2Bcos%28x%29%2F%281-cos%28x%29%29
Actually it's a good sign that the identities to use were not obvious. There are none that directly connect cos and csc.

But there are identities that connect cos to sin and sin to csc. So that is the path we will follow: Change cos to sin and then sin to csc. The identity that connects sin(x) and cos(x) is:
sin%5E2%28x%29%2Bcos%5E2%28x%29+=+1
(or its "cousins" cos%5E2%28x%29+=+1+-+sin%5E2%28x%29 and sin%5E2%28x%29+=+1-cos%5E2%28x%29). In order to use these we will need to have cos%5E2%28x%29. So that is our first task: Manipulate the expression so that the cos(x)'s turn into cos%5E2%28x%29. This might be the hardest part of this problem.

When I started looking at this I looked at the denominators and saw the first one as "a+b" and the second one as "a-b". And I know well that %28a%2Bb%29%28a-b%29+=+a%5E2-b%5E2. So I saw that that a%5E2-b%5E2 will have a


Adding...
%281%2Bcos%5E2%28x%29%29%2F%281-cos%5E2%28x%29%29
And isn't it nice that the cos(x)'s in the numerator canceled out!

Now we can substitute in to get sin's:
%281%2B%281-sin%5E2%28x%29%29%29%2Fsin%5E2%28x%29
which simplifies to:
%282-sin%5E2%28x%29%29%2Fsin%5E2%28x%29

Now we can substitute in to get the csc's:
%282-1%2Fcsc%5E2%28x%29%29%2F%281%2Fcsc%5E2%28x%29%29
All that's left is to simplify:

%282csc%5E2%28x%29-1%29%2F1
which simplifies to:
2csc%5E2%28x%29-1