How do you evaluate $tan^-1(tan((13pi)/6))$ ?

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Answer 1 · 2017-01-09T12:22:17

$pi/6$ .

$tan^-1{tan((13pi)/6)}=tan^-1{tan((12pi+pi)/6)}$

$=tan^-1{tan(2pi+(pi)/6)}$

$=tan^-1{tan((pi)/6)}...[because, tan(2pi+x)=tanx]$

$=pi/6$

In the last step, we have used the following Defn. of $tan^-1$ fun.:

$theta=tan^-1x, x in RR iff tantheta=x, theta in (-pi/2,pi/2).$

How do you evaluate tan^-1(tan((13pi)/6))tan^-1(tan((13pi)/6))?