How do you evaluate #tan^-1(tan((19pi)/10))#?

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Answer 1 · 2016-07-21T12:17:06

#(19/10)pi#.

#tan^(-1)tan((19pi)/10)# is

the angle whose tangent is #tan ((19pi)/10)#

#=(19/10)pi#.

From the definitions of successive operations

#f f^(-1) (y)=y# and

#f^(-1)f(x)=x,#

only the operand value has to be given as the answer and

not any other value, from the general value

#npi+(19pi)/10, n = 0, +-1, +-2, +-3....#

Here, #f = tan and f^(-1)=tan^(-1)#.