SOLUTION: Prove that "1+sin^2x=csc^2x-cot^2x+sin^2x"

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Question 594379: Prove that "1+sin^2x=csc^2x-cot^2x+sin^2x"
Found 2 solutions by lwsshak3, jsmallt9:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Prove that "1+sin^2x=csc^2x-cot^2x+sin^2x"
Starting with right side:
csc^2x-cot^2x+sin^2x
=1/sin^2-cos^2/sin^2+sin^2
=1-cos^2/sin^2+sin^
=sin^2/sin^2+sin^2
=1+sin^2
verified:
right side=left side

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
1%2Bsin%5E2%28x%29=csc%5E2%28x%29-cot%5E2%28x%29%2Bsin%5E2%28x%29
There are many techniques to proving identities (equations that are true for all x's. One technique is to match terms. We already have a match (which I have highlighted in red):
1%2Bred%28sin%5E2%28x%29%29=csc%5E2%28x%29-cot%5E2%28x%29%2Bred%28sin%5E2%28x%29%29
We will leave the matched terms alone. Now we must find a way to show that the remaining terms on each side are equal. The unmatched terms are 1, csc%5E2%28x%29 and cot%5E2%28x%29. If we know our properties well we know that one of the Pythagorean properties has exactly those three terms:
1+%2B+cot%5E2%28x%29+=+csc%5E2%28x%29. If we subtract cot%5E2%28x%29 from each side we get:
1+=+csc%5E2%28x%29+-+cot%5E2%28x%29
So we can substitute the right side for the 1 in your equation giving us:

And we are finished.