How do you evaluate ${tan}^{- 1} (tan (\frac{7 π}{6}))$ $tan^-1(tan((7pi)/6))$ ?

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Answer 1 · 2016-07-24T09:31:55

$\frac{7 π}{6}$ $(7pi)/6$

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

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

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

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

So, the answer is $\frac{7 π}{6}$ $(7pi)/6$

The conventional restriction on

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

has no relevance for this double operation.

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

the value will be $(\frac{π}{6})$ $(pi/6)$ .

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

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

The source for the value $\frac{7 π}{6}$ $(7pi)/6$ is the general value

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

..

How do you evaluate tan^-1(tan((7pi)/6))tan−1(tan(7π6))tan^-1(tan((7pi)/6))?