How do you find #tan(cos^-1(3/x))#?

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Answer 1 · 2016-09-20T03:00:54

#+-sqrt(x^2-9)/3, |x|>=3#.

#Choose - sign, for #x <-3#. .

Let #a = cos^(-1)(3/x) in [0, pi]#, for principal value. Then,

#cos a = 3/x, |x|>=3#.# sin a >=0#, for a in #Q_1 and Q_2# and

tan a and cos a have the same sign.

The given expression is

#tan a = sin a/cos a= +-sqrt(1-9/x^2)/(3/x)=+-sqrt(x^2-9)/3#.

Choose #-# sign, for #x < -3# and + sign for #x>3#..

Note that, when #x = -3, a = pi# and when x = 3, a = 0. .