How do you verify $cot (\frac{x}{2}) = \frac{tan x}{sec x - 1}$ $cot (x/2) = tan x/(sec x-1)$ ?

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How do you verify $cot (\frac{x}{2}) = \frac{tan x}{sec x - 1}$ $cot (x/2) = tan x/(sec x-1)$ ?

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Answer 1 · 2015-05-21T04:39:20

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

$= \frac{2 sin (\frac{x}{2}) . cos (\frac{x}{2})}{2 {sin}^{2} (\frac{x}{2})} = \frac{cos (\frac{x}{2})}{sin (\frac{x}{2})} = cot (\frac{x}{2})$ $= (2sin (x/2).cos (x/2))/(2sin^2 (x/2)) = cos (x/2)/sin (x/2) = cot (x/2)$

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How do you verify $cot (\frac{x}{2}) = \frac{tan x}{sec x - 1}$ $cot (x/2) = tan x/(sec x-1)$ ?

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How do you verify $cot (\frac{x}{2}) = \frac{tan x}{sec x - 1}$ $cot (x/2) = tan x/(sec x-1)$ ?