How do you evaluate $tan^-1(tan((19pi)/10))$ ?

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Answer 1 · 2016-07-21T12:17:06

$(19/10)pi$ .

$tan^(-1)tan((19pi)/10)$ is

the angle whose tangent is $tan ((19pi)/10)$

$=(19/10)pi$ .

From the definitions of successive operations

$f f^(-1) (y)=y$ and

$f^(-1)f(x)=x,$

only the operand value has to be given as the answer and

not any other value, from the general value

$npi+(19pi)/10, n = 0, +-1, +-2, +-3....$

Here, $f = tan and f^(-1)=tan^(-1)$ .

How do you evaluate tan^-1(tan((19pi)/10))tan^-1(tan((19pi)/10))?