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

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Answer 1 · 2016-11-12T08:51:31

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${(csc x - cot x)}^{2} = \frac{1 - cos x}{1 + cos x}$ $(cscx-cotx)^2=(1-cosx)/(1+cosx)$

Left Side: $= {(csc x - cot x)}^{2}$ $=(cscx-cotx)^2$

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

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

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

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

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

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

$=(cancel(1-cosx)(1-cosx))/(cancel(1-cosx)(1+cosx))$

$=(1-cosx)/(1+cosx)$

$:.=$ Right Side

How do you verify (cscx-cotx)^2=(1-cosx)/(1+cosx)(cscx−cotx)2=1−cosx1+cosx(cscx-cotx)^2=(1-cosx)/(1+cosx)?