# How do you find the remaining trigonometric ratios if tan(alpha) = 9 and 0 < alpha < pi/2?

Jul 20, 2015

Use SOH-CAH-TOA mnemonic and Pythagoras to find:

$\sin \left(\alpha\right) = \frac{9}{\sqrt{82}}$ and $\cos \left(\alpha\right) = \frac{1}{\sqrt{82}}$

$\csc \left(\alpha\right) = \frac{\sqrt{82}}{9}$, $\sec \left(\alpha\right) = \sqrt{82}$ and $\cot \left(\alpha\right) = \frac{1}{9}$

#### Explanation:

$\tan \left(\alpha\right) = 9$ is the $\text{opposite"/"adjacent}$ ratio of a right angled triangle with angle $\alpha$.

For our purposes, it does not matter what size the triangle is - just its proportions. So let the length of the opposite side be $9$ and the length of the adjacent side be $1$. Then the length of the hypotenuse is $\sqrt{{9}^{2} + {1}^{2}} = \sqrt{82}$.

Hence $\sin \left(\alpha\right) = \text{opposite"/"hypotenuse} = \frac{9}{\sqrt{82}}$

and $\cos \left(\alpha\right) = \text{adjacent"/"hypotenuse} = \frac{1}{\sqrt{82}}$

Jul 20, 2015

Find other trig functions knowing tan x = 9

#### Explanation:

Use a calculator.
tan x = 9 --> x = 83.66 deg
sin 83.66 = 0.99
cos 83.66 = 0.11
$\cot x = \frac{1}{9}$
$\sec x = \frac{1}{0.11} = 9.09$
$\csc x = \frac{1}{0.99} = 1.01$