# How do you find the value of the other five trigonometric functions, given tan x=4/11, sec x< 0?

Oct 10, 2015

#### Answer:

$\sin x = - \frac{2}{\sqrt{15}} , \cos x = - \sqrt{\frac{11}{15}}$

$\tan x = \frac{4}{11} , \cot x = \frac{11}{4}$

$\sec x = - \sqrt{\frac{15}{11}} , \csc x = - \frac{\sqrt{15}}{2}$

#### Explanation:

${\sec}^{2} x = {\tan}^{2} x + 1$

${\sec}^{2} x = \frac{4}{11} + 1 = \frac{4}{11} + \frac{11}{11} = \frac{15}{11}$

$\sec x = \sqrt{\frac{15}{11}} \vee \sec x = - \sqrt{\frac{15}{11}}$

So, $\sec x = - \sqrt{\frac{15}{11}}$

$\cos x = \frac{1}{\sec} x$

$\cos x = - \sqrt{\frac{11}{15}}$

${\sin}^{2} x + {\cos}^{2} x = 1 \implies {\sin}^{2} x = 1 - {\cos}^{2} x$

${\sin}^{2} x = 1 - \frac{11}{15} = \frac{4}{15}$

$\sin x = - \frac{2}{\sqrt{15}} \vee \sin x = \frac{2}{\sqrt{15}}$

Since $\tan x = \frac{4}{11} = \sin \frac{x}{\cos} x$ and $\cos x < 0$ it must hold $\sin x < 0$

So, $\sin x = - \frac{2}{\sqrt{15}}$

$\cot x = \frac{1}{\tan} x = \frac{11}{4}$

$\csc x = \frac{1}{\sin} x = - \frac{\sqrt{15}}{2}$

Finally:

$\sin x = - \frac{2}{\sqrt{15}} , \cos x = - \sqrt{\frac{11}{15}}$

$\tan x = \frac{4}{11} , \cot x = \frac{11}{4}$

$\sec x = - \sqrt{\frac{15}{11}} , \csc x = - \frac{\sqrt{15}}{2}$