Question #49284

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Question #49284

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Answer 1 · 2016-11-17T18:10:24

$1 - (\frac{{sin}^{2} x}{1 + cos x} + \frac{1 + cos x}{sin x} - \frac{sin x}{1 - cos X})$ $1- (sin^2x/(1+cosx) + (1+cosx)/sinx - sinx/(1-cosX))$

$= 1 - (\frac{(1 - cos x) (1 + cos x)}{1 + cos x} + \frac{1 + cos x}{sin x} - \frac{sin x (1 + cos x)}{(1 - cos x) (1 + cos x)})$ $=1-(((1-cosx)(1+cosx))/(1+cosx)+(1+cosx)/sinx-(sinx(1+cosx))/((1-cosx)(1+cosx)))$

$= 1 - 1 + cos x - \frac{1 + cos x}{sin x} + \frac{sin x (1 + cos x)}{1 - {cos}^{2} x}$ $=1-1+cosx-(1+cosx)/sinx +(sinx(1+cosx))/(1-cos^2x)$

$= cos x - \frac{1 + cos x}{sin x} + \frac{sin x (1 + cos x)}{{sin}^{2} x}$ $=cosx-(1+cosx)/sinx +(sinx(1+cosx))/sin^2x$

$= cos x - \frac{1 + cos x}{sin x} + \frac{1 + cos x}{sin x}$ $=cosx-(1+cosx)/sinx +(1+cosx)/sinx$

$= cos x$ $=cosx$

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Question #49284

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