How do you prove $tan (\frac{x}{2}) = sin x + cos x cot x - cot x$ $tan(x/2)= sinx+cosxcotx-cotx$ ?

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How do you prove $tan (\frac{x}{2}) = sin x + cos x cot x - cot x$ $tan(x/2)= sinx+cosxcotx-cotx$ ?

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Answer 1 · 2015-12-21T16:23:07

Develop the right side.

Explanation:

We know that $tan (\frac{x}{2}) = \frac{1 - cos (x)}{sin (x)}$ $tan(x/2) = (1 - cos(x))/sin(x)$ . So we develop the right side of the equality. $cot (x) = \frac{1}{tan (x)}$ $cot(x) = 1/tan(x)$ so :

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

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How do you prove $tan (\frac{x}{2}) = sin x + cos x cot x - cot x$ $tan(x/2)= sinx+cosxcotx-cotx$ ?

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Explanation:

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How do you prove $tan (\frac{x}{2}) = sin x + cos x cot x - cot x$ $tan(x/2)= sinx+cosxcotx-cotx$ ?