How do you evaluate ${tan}^{- 1} (tan (\frac{19 π}{10}))$ $tan^-1(tan((19pi)/10))$ ?

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Answer 1 · 2016-07-21T12:17:06

$(\frac{19}{10}) π$ $(19/10)pi$ .

${tan}^{- 1} tan (\frac{19 π}{10})$ $tan^(-1)tan((19pi)/10)$ is

the angle whose tangent is $tan (\frac{19 π}{10})$ $tan ((19pi)/10)$

$= (\frac{19}{10}) π$ $=(19/10)pi$ .

From the definitions of successive operations

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

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

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(19π10))tan^-1(tan((19pi)/10))?