How do you find the exact value of ${cot}^{- 1} (- \sqrt{3})$ $cot^-1(-sqrt3)$ ?

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Answer 1 · 2017-06-11T10:08:29

${cot}^{- 1} (- \sqrt{3}) = - \frac{π}{6}$ $cot^(-1)(-sqrt3)=-pi/6$

The range for ${cot}^{- 1} x$ $cot^(-1)x$ is from $- \frac{π}{2}$ $-pi/2$ to $\frac{π}{2}$ $pi/2$

As $cot (- \frac{π}{6}) = - \sqrt{3}$ $cot(-pi/6)=-sqrt3$

${cot}^{- 1} (- \sqrt{3}) = - \frac{π}{6}$ $cot^(-1)(-sqrt3)=-pi/6$

How do you find the exact value of cot−1(−√3)cot^-1(-sqrt3)?