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

Question

7633 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-24T09:31:55

$(7pi)/6$

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

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

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

Here, $f = tan, f^(-1) = tan^(-1)$ and the operand $a = (7pi)/6$

So, the answer is $(7pi)/6$

The conventional restriction on

'a' as the principal value $in [-pi/2, pi/2]$

has no relevance for this double operation.

If the question is about $tan^(-1) tan (pi/6)$ ,

the value will be $(pi/6)$ .

In either case the tan value = $1/sqrt 3$ ..

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

The source for the value $(7pi)/6$ is the general value

$npi+pi/6, n = 1$ .. . .

..

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