How do you verify $\frac{sin x}{1 - cos x} + \frac{1 - cos x}{sin x} = 2 csc x$ $sinx/(1-cosx) + (1-cosx)/sinx = 2cscx$ ?

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Answer 1 · 2018-03-06T20:51:10

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$L H S : \frac{sin x}{1 - cos x} + \frac{1 - cos x}{sin x}$ $LHS: sin x/(1-cos x) +(1-cosx)/sin x$

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

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

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

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

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

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

$=(2(cancel(1-cosx)))/(sinx cancel((1-cosx)))$

$=2/sinx$

$=2*1/sinx$

$=2 csc x$

$=RHS$

How do you verify sinx/(1-cosx) + (1-cosx)/sinx = 2cscxsinx1−cosx+1−cosxsinx=2cscxsinx/(1-cosx) + (1-cosx)/sinx = 2cscx?