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. .

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