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

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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#.. . .

..