Question 1004983: Alright this is a trignometry identities question.
The equation is tan^2x+1/tan^2x-1 = 1/sin^2x-cos^2x
Now I need to make the left side equal the right side.
I know that tan^2x+1 is sec^2x or 1/cos^2x and the bottom tan changes to sin^2x/cos^2x -1 but I'm not sure what to do after that because I have a fraction over a fraction.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! it's actually easier to solve than you might think.
first recognize that 1 is equivalent to sin^2(x) + cos^2(x).
also recognize that tan^2(x) = sin^2(x) / cos^2(x).
you can either multiply numerator and denominator of the expression on the left by cos^2(x), or you can divide numerator and denominator of the expression on the right by cos^2(x).
working on the left:
your numerator of (tan^2(x) + 1) * cos^2(x) becomes:
tan^2(x) * cos^2(x) + 1 * cos^2(x) which becomes:
sin^2(x) + cos^2(x).
similarly, your denominator becomes sin^2(x) - cos^2(x).
working on the right:
your numerator of 1 becomes sin^2(x) + cos^(x).
(sin^2(x) + cos^2(x)) / cos^2(x) becomes:
sin^2(x) / cos^2(x) + cos^2(x) / cos^2(x) which becomes:
tan^2(x) + 1
similarly, your denominator becomes tan^2(x) - 1.
you are correct that tan^2(x) + 1 = sec^2(x), but that didn't really help you here, as far as i can tell.
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