# How do you evaluate tan^-1(tan((7pi)/6))?

Jul 24, 2016

$\frac{7 \pi}{6}$

#### Explanation:

When the operation is "inverse of a function over the function# on

an operand 'a', the result is 'a'.

Symbolically., ${f}^{- 1} f \left(a\right) = a$.

Here, $f = \tan , {f}^{- 1} = {\tan}^{- 1}$ and the operand $a = \frac{7 \pi}{6}$

So, the answer is $\frac{7 \pi}{6}$

The conventional restriction on

'a' as the principal value $\in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

has no relevance for this double operation.

If the question is about ${\tan}^{- 1} \tan \left(\frac{\pi}{6}\right)$,

the value will be $\left(\frac{\pi}{6}\right)$.

In either case the tan value = $\frac{1}{\sqrt{3}}$..

Of course, the principal value of ${\tan}^{- 1} \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

The source for the value $\frac{7 \pi}{6}$ is the general value

$n \pi + \frac{\pi}{6} , n = 1$.. . .

..