How do you prove that $tan 15 = 2 - \sqrt{3}$ $tan15=2-sqrt3$ ?

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How do you prove that $tan 15 = 2 - \sqrt{3}$ $tan15=2-sqrt3$ ?

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Answer 1 · 2018-05-26T04:26:23

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Explanation:

RTP: $tan 15 = 2 - \sqrt{3}$ $tan15=2-sqrt3$

$tan 15 = tan (45 - 30)$ $tan15 = tan(45-30)$

Recall: $tan (a - b) = \frac{tan a - tan b}{1 + tan a tan b}$ $tan(a-b)=(tana-tanb)/(1+tanatanb)$

$tan (45 - 30) = \frac{tan 45 - tan 30}{1 + tan 45 tan 30}$ $tan(45-30)=(tan45-tan30)/(1+tan45tan30)$

= $\frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$ $(1-1/sqrt3)/(1+1/sqrt3)$

= $\frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}$ $((sqrt3-1)/sqrt3)/((sqrt3+1)/sqrt3)$

= $(\frac{\sqrt{3} - 1}{\sqrt{3}}) \times (\frac{\sqrt{3}}{\sqrt{3} + 1})$ $((sqrt3-1)/sqrt3)times(sqrt3/(sqrt3+1))$

= $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ $(sqrt3-1)/(sqrt3+1)$

Rationalise the denominator
= $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$ $(sqrt3-1)/(sqrt3+1)times(sqrt3-1)/(sqrt3-1)$

= $\frac{3 - 2 \sqrt{3} + 1}{3 - 1}$ $(3-2sqrt3+1)/(3-1)$

= $\frac{4 - 2 \sqrt{3}}{2}$ $(4-2sqrt3)/2$

Take out the common factor
= $\frac{2 (2 - \sqrt{3})}{2}$ $(2(2-sqrt3))/2$

Simplify
= $2 - \sqrt{3}$ $2-sqrt3$

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How do you prove that $tan 15 = 2 - \sqrt{3}$ $tan15=2-sqrt3$ ?

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