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

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

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Answer 1 · 2015-04-15T15:28:32

In this way.

The first member is:

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

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

that is the second member.

Answer 2 · 2015-04-15T15:37:22

${(csc x + cot x)}^{2}$ $(csc x + cotx)^2$

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

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

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

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

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

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

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