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sin(2A)/[1-(sinA)^2]=2tan(A) can anyone start me out??
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ok.. you need to know a couple of things to start this problem 1. sin(2A) = 2sin(A)cos(A) 2. 1 - sin^2(A) = cos^2(A) .... or cos(A)cos(A) 3. \[\tan(A) = \frac{\sin(A)}{\cos(A)}\] then you fraction becomes \[\frac{2\sin(A)\cos(A)}{\cos(A)\cos(A)}\] eliminate the common factor and you should have your answer
oh my goodness! thank you so much!
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