How do you prove that ${cot}^{2} x = \frac{1 + cos 2 x}{1 - cos 2 x}$ $cot^2x = (1+cos2x)/(1-cos2x)$ ?

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How do you prove that ${cot}^{2} x = \frac{1 + cos 2 x}{1 - cos 2 x}$ $cot^2x = (1+cos2x)/(1-cos2x)$ ?

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Answer 1 · 2015-12-06T15:23:31

$cos 2 x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x}$ $cos2x= (1-tan^2x)/(1+tan^2x)$

$\Rightarrow \frac{1 + cos 2 x}{1 - cos 2 x} = \frac{1 + (\frac{1 - {tan}^{2} x}{1 + {tan}^{2} x})}{1 - (\frac{1 - {tan}^{2} x}{1 + {tan}^{2} x})}$ $=> (1+cos2x)/(1-cos2x) = (1+((1-tan^2x)/(1+tan^2x)) )/(1-((1-tan^2x)/(1+tan^2x)))$

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

$= \frac{2}{2 {tan}^{2} x} = {cot}^{2} x$ $= 2/(2tan^2x)=cot^2x$

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How do you prove that ${cot}^{2} x = \frac{1 + cos 2 x}{1 - cos 2 x}$ $cot^2x = (1+cos2x)/(1-cos2x)$ ?

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How do you prove that cot2x=1+cos2x1−cos2x cot^2x = (1+cos2x)/(1-cos2x) ?

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How do you prove that ${cot}^{2} x = \frac{1 + cos 2 x}{1 - cos 2 x}$ $cot^2x = (1+cos2x)/(1-cos2x)$ ?