# How do you find the exact value of tan[arc cos(-1/3)]?

Jul 28, 2015

You use the trigonometric Identity $\tan \left(\theta\right) = \sqrt{\left(\frac{1}{\cos} ^ 2 \left(\theta\right) - 1\right)}$

Result : $\tan \left[\arccos \left(- \frac{1}{3}\right)\right] = \textcolor{b l u e}{2 \sqrt{2}}$

#### Explanation:

Start by letting $\arccos \left(- \frac{1}{3}\right)$ to be an angle $\theta$

$\implies \arccos \left(- \frac{1}{3}\right) = \theta$

$\implies \cos \left(\theta\right) = - \frac{1}{3}$

This means that we are now looking for $\tan \left(\theta\right)$

Next, use the identity : ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$

Divide all both sides by ${\cos}^{2} \left(\theta\right)$ to have,

$1 + {\tan}^{2} \left(\theta\right) = \frac{1}{\cos} ^ 2 \left(\theta\right)$

$\implies {\tan}^{2} \left(\theta\right) = \frac{1}{\cos} ^ 2 \left(\theta\right) - 1$

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

Recall, we said earlier that $\cos \left(\theta\right) = - \frac{1}{3}$

$\implies \tan \left(\theta\right) = \sqrt{\frac{1}{- \frac{1}{3}} ^ 2 - 1} = \sqrt{\frac{1}{\frac{1}{9}} - 1} = \sqrt{9 - 1} = \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = \textcolor{b l u e}{2 \sqrt{2}}$