# How do you determine all trigonometric ratios given secθ=6/5?

Nov 17, 2015

$\sin \theta = \frac{\sqrt{11}}{6} , \tan \theta = \frac{\sqrt{11}}{5} , \cos \theta = \frac{5}{6} ,$
$\csc \theta = \frac{6 \sqrt{11}}{11} , \cot \theta = \frac{5 \sqrt{11}}{11}$

#### Explanation:

If $\sec \theta = \frac{6}{5}$
$\cos \theta = \frac{1}{\sec \theta} = \frac{5}{6}$

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$
$\sin \theta = \sqrt{1 - {\cos}^{2} \theta} = \sqrt{1 - {\left(\frac{5}{6}\right)}^{2}} = \frac{\sqrt{11}}{6}$

$\csc \theta = \frac{1}{\sin} \theta = \frac{6}{\sqrt{11}} = \frac{6 \sqrt{11}}{11}$

$1 + {\tan}^{2} \theta = {\sec}^{2} \theta$
$\tan \theta = \sqrt{{\sec}^{2} \theta - 1} = \sqrt{{\left(\frac{6}{5}\right)}^{2} - 1} = \frac{\sqrt{11}}{5}$

$\cot \theta = \frac{1}{\tan} \theta = \frac{5}{\sqrt{11}} = \frac{5 \sqrt{11}}{11}$