How do you evaluate ${tan}^{- 1} (tan (\frac{5 π}{6}))$ $tan^-1(tan((5pi)/6))$ ?

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Answer 1 · 2016-08-04T17:38:22

$= 5 \frac{π}{6}$ $=5pi/6$

${tan}^{- 1} (tan (5 \frac{π}{6}))$ $tan^-1(tan(5pi/6))$
$= 5 \frac{π}{6}$ $=5pi/6$

Answer 2 · 2016-08-04T20:00:48

${tan}^{- 1} (tan (\frac{5 π}{6})) = - \frac{π}{6}$ $tan^(-1)(tan((5pi)/6)) = -pi/6$

$θ = {tan}^{- 1} (tan (\frac{5 π}{6}))$ $theta = tan^(-1)(tan((5pi)/6))$ by definition satisfies both of the conditions:

Note that $tan$ $tan$ has period $π$ $pi$ , so for any integer $n$ $n$ :

$tan (\frac{5 π}{6} + n π) = tan (\frac{5 π}{6})$ $tan ((5pi)/6 + npi) = tan ((5pi)/6)$

When $n = - 1$ $n = -1$ , we have:

$\frac{5 π}{6} + n π = \frac{5 π}{6} - π = - \frac{π}{6}$ $(5pi)/6+npi = (5pi)/6 - pi = -pi/6$

which lies in the range $(- \frac{π}{2}, \frac{π}{2})$ $(-pi/2, pi/2)$ , so satisfies the second condition for ${tan}^{- 1}$ $tan^(-1)$

Thus:

${tan}^{- 1} (tan (\frac{5 π}{6})) = - \frac{π}{6}$ $tan^(-1)(tan((5pi)/6)) = -pi/6$

How do you evaluate tan−1(tan(5π6))tan^-1(tan((5pi)/6))?