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

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Answer 1 · 2018-02-25T16:16:44

A fundamental property of all inverses is:

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

Applying the above property:

${tan}^{- 1} (tan (x)) = x$ $tan^-1(tan(x)) = x$

Substitute $x = - \frac{π}{10}$ $x = -pi/10$ :

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

How do you evaluate tan−1(tan(−π10))tan^-1(tan(-(pi)/10))?