How do you evaluate $cot (\frac{17 π}{6})$ $cot ((17pi)/6)$ ?

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How do you evaluate $cot (\frac{17 π}{6})$ $cot ((17pi)/6)$ ?

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Answer 1 · 2016-03-18T17:34:03

It is

$cot (17 \cdot \frac{π}{6}) = cot (\frac{12 π + 5 π}{6}) = cot (2 π + 5 \frac{π}{6}) = cot (5 \frac{π}{6}) = = cot (π - \frac{π}{6}) = - cot (\frac{π}{6}) = - \sqrt{3}$ $cot(17*pi/6)=cot([12pi+5pi]/6)=cot(2pi+5pi/6)=cot(5pi/6)= =cot(pi-pi/6)=-cot(pi/6)=-sqrt3$

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How do you evaluate $cot (\frac{17 π}{6})$ $cot ((17pi)/6)$ ?

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How do you evaluate $cot (\frac{17 π}{6})$ $cot ((17pi)/6)$ ?