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simplify: (sin^2x)/(sec^2x-1)
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help :(
sin^2 + cos^2 = 1 dividing through by cos^2 tan^2 + 1 = sec^2 therefore sec^2 -1 = tan^2
so the question becomes\[\frac{\sin^2(x)}{\tan^2(x)}\]
\[= \sin^2(x) \times \frac{\cos^2(x)}{\sin^2(x)}\]
\[= \cos^2(x)\]
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remember that \[\tan(x) = \frac{\sin(x)}{\cos(x)}\] then when divide by something it is the same as flipping that"something", and multiplying
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