# How do you find the exact value of the five remaining trigonometric function given sectheta=-sqrt10 and the angle is in standard position in quadrant III?

Dec 26, 2016

$\cos \theta = - \frac{1}{\sqrt{10}}$;
$\sin \theta = - \frac{3}{\sqrt{10}}$;
$\tan \theta = 3$

$\cot \theta = \frac{1}{3}$

$\csc \theta = - \frac{\sqrt{10}}{3}$

#### Explanation:

Since

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

then

$\cos \theta = \frac{1}{-} \sqrt{10} = - \frac{1}{\sqrt{10}}$

Since the angle is in standard position in quadrant III, and

$\sin \theta = \pm \sqrt{1 - {\cos}^{2} \theta}$

then $\sin \theta = \textcolor{red}{-} \sqrt{1 - {\left(\frac{1}{- \sqrt{10}}\right)}^{2}} = - \sqrt{1 - \frac{1}{10}} = - \sqrt{\frac{9}{10}} = - \frac{3}{\sqrt{10}}$

So

$\tan \theta = \sin \frac{\theta}{\cos} \theta = \frac{- \frac{3}{\cancel{\sqrt{10}}}}{-} \left(\frac{1}{\cancel{\sqrt{10}}}\right) = 3$

$\cot \theta = \frac{1}{\tan} \theta = \frac{1}{3}$

$\csc \theta = \frac{1}{\sin} \theta = \frac{1}{-} \sqrt{\frac{9}{10}} = - \frac{\sqrt{10}}{3}$