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

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Answer 1 · 2016-08-04T17:38:22

$=5pi/6$

$tan^-1(tan(5pi/6))$
$=5pi/6$

Answer 2 · 2016-08-04T20:00:48

$tan^(-1)(tan((5pi)/6)) = -pi/6$

$theta = tan^(-1)(tan((5pi)/6))$ by definition satisfies both of the conditions:

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

$tan ((5pi)/6 + npi) = tan ((5pi)/6)$

When $n = -1$ , we have:

$(5pi)/6+npi = (5pi)/6 - pi = -pi/6$

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

Thus:

$tan^(-1)(tan((5pi)/6)) = -pi/6$

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