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

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

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Answer 1 · 2015-12-08T05:44:24

Prove trig expression

Explanation:

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

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

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How do you prove (cscx−cotx)2=1−cosx1+cosx(csc x - cot x)^2 = (1 - cos x)/(1+cosx)?

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