How do you evaluate $cot (\frac{π}{2} + 11 \frac{π}{6})$ $cot (pi/2 + 11pi/6)$ ?

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How do you evaluate $cot (\frac{π}{2} + 11 \frac{π}{6})$ $cot (pi/2 + 11pi/6)$ ?

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Answer 1 · 2016-07-04T07:46:32

$cot (\frac{π}{2} + \frac{11 π}{6}) = \frac{1}{\sqrt{3}}$ $cot(pi/2+(11pi)/6)=1/sqrt3$

Explanation:

It is observed that $\frac{11 π}{6}$ $(11pi)/6$ is just less than $2 π$ $2pi$ and all trigonometric ratios get repeated after $2 π$ $2pi$ , Hence, we can have

$cot (\frac{π}{2} + \frac{11 π}{6})$ $cot(pi/2+(11pi)/6)$

= $cot (\frac{π}{2} + \frac{12 π}{6} - \frac{π}{6})$ $cot(pi/2+(12pi)/6-pi/6)$

= $cot (\frac{π}{2} + 2 π - \frac{π}{6})$ $cot(pi/2+2pi-pi/6)$

= $cot (\frac{π}{2} - \frac{π}{6})$ $cot(pi/2-pi/6)$

Further as cotangent of a complementary angle and its tangent are equal, we have

$cot (\frac{π}{2} - \frac{π}{6})$ $cot(pi/2-pi/6)$

= $\frac{tan π}{6}$ $tanpi/6$

= $\frac{1}{\sqrt{3}}$ $1/sqrt3$

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How do you evaluate $cot (\frac{π}{2} + 11 \frac{π}{6})$ $cot (pi/2 + 11pi/6)$ ?

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