# How do you find the value of tan(csc^-1(2))?

Apr 24, 2018

$\sqrt{3}$

#### Explanation:

First you need to find the angle that corresponds to a csc value of 2, which is what inverse csc is doing.
${\csc}^{-} 1 = \frac{1}{{\cos}^{-} 1} = 2$
${\cos}^{-} 1 = \frac{1}{2}$
So find when cos is 1/2, which is at $\frac{\pi}{3}$
Then plug in $\frac{\pi}{3}$ for ${\csc}^{-} 1 \left(2\right)$
Evaluate $\tan \left(\frac{\pi}{3}\right)$
$\tan \left(\frac{\pi}{3}\right) = \sqrt{3}$

Apr 24, 2018

Use the identity:

$\tan \left({\csc}^{-} 1 \left(x\right)\right) = \frac{1}{\sqrt{{x}^{2} - 1}}$

#### Explanation:

Please see this reference section Relationships between trigonometric functions and inverse trigonometric functions. I am referring you to this section because it contains a table that will help you if your current studies require you to do many problems of this type.

The table gives the following identity:

$\tan \left({\csc}^{-} 1 \left(x\right)\right) = \frac{1}{\sqrt{{x}^{2} - 1}}$

Please notice that there is a nice triangle drawing to the right within the table:

Substitute $x = 2$ into the identity:

$\tan \left({\csc}^{-} 1 \left(2\right)\right) = \frac{1}{\sqrt{{2}^{2} - 1}}$

$\tan \left({\csc}^{-} 1 \left(2\right)\right) = \frac{1}{\sqrt{4 - 1}}$

$\tan \left({\csc}^{-} 1 \left(2\right)\right) = \frac{1}{\sqrt{3}}$

Rationalize the denominator:

$\tan \left({\csc}^{-} 1 \left(2\right)\right) = \frac{\sqrt{3}}{3}$