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

Sep 20, 2016

$\pm \frac{\sqrt{{x}^{2} - 9}}{3} , | x | \ge 3$.

$C h \infty s e - s i g n , f \mathmr{and}$x <-3#. .

#### Explanation:

Let $a = {\cos}^{- 1} \left(\frac{3}{x}\right) \in \left[0 , \pi\right]$, for principal value. Then,

$\cos a = \frac{3}{x} , | x | \ge 3$.$\sin a \ge 0$, for a in ${Q}_{1} \mathmr{and} {Q}_{2}$ and

tan a and cos a have the same sign.

The given expression is

$\tan a = \sin \frac{a}{\cos} a = \pm \frac{\sqrt{1 - \frac{9}{x} ^ 2}}{\frac{3}{x}} = \pm \frac{\sqrt{{x}^{2} - 9}}{3}$.

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

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