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Rewrite the expression (cos^2x)/(1+sinx) so that there is no fraction.
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\[\frac{1-\sin^2x}{1+\sin^2x}\] Remember difference of squares? \[\frac{(1-\sin x)(1+\sin x)}{1+\sin x}\] can you solve now?
is it 1/sinx = cscx?
1+sin x cancels out from both numerator and denominator and so leaving you behind with 1-sin x
Got it! Thanks!
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