How do you find the exact value of tan (23pi/6)?

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How do you find the exact value of tan (23pi/6)?

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Answer 1 · 2018-03-13T05:39:26

$- \frac{1}{\sqrt{3}}, or, - \frac{\sqrt{3}}{3}$ $-1/sqrt3, or, -sqrt3/3$ .

Explanation:

$tan (\frac{23}{6} π) = tan {\frac{24 - 1}{6} π} = tan {(\frac{24}{6} - \frac{1}{6}) π}$ $tan(23/6pi)=tan{(24-1)/6pi}=tan{(24/6-1/6)pi}$ ,

$= tan (4 π - \frac{π}{6})$ $=tan(4pi-pi/6)$ .

Since, $(4 π - \frac{π}{6})$ $(4pi-pi/6)$ lies in the fourth quadrant, where, tan is

-ve, we get,

$tan (\frac{23}{6} π) = tan (4 π - \frac{π}{6}) = - tan (\frac{π}{6}) = - \frac{1}{\sqrt{3}}, or, - \frac{\sqrt{3}}{3}$ $tan(23/6pi)=tan(4pi-pi/6)=-tan(pi/6)=-1/sqrt3, or, -sqrt3/3$ .

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How do you find the exact value of tan (23pi/6)?

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