Question #50e6d

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Answer 1 · 2016-05-04T18:41:07

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

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

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

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

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

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

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

$= 2 \frac{1}{sin x}$ $=2 1/sinx$

$= 2 csc x$ $=2csc x$

$=$ $=$ Right Side