# How do you use fundamental identities to find the values of the trigonometric values given θ be an angle in quadrant III, such that cot θ = (√11)/(5)?

The given $\cot \theta = \frac{\sqrt{11}}{5}$,
the answers $\tan \theta = \frac{5}{\sqrt{11}}$, $\text{ } \sin \theta = - \frac{5}{6}$, $\text{ } \cos \theta = - \frac{\sqrt{11}}{6}$,$\text{ } \csc \theta = - \frac{6}{5}$, $\text{ } \sec \theta = \frac{- 6 \sqrt{11}}{11}$

#### Explanation:

the given $\cot \theta$ at the third quadrant x is negative and y is negative.

The formulas:
$\tan \theta = \frac{1}{\cot} \theta$
$\csc \theta = \pm \sqrt{{\cot}^{2} \theta + 1}$
$\sin \theta = \frac{1}{\csc} \theta = \frac{1}{\pm \sqrt{{\cot}^{2} \theta + 1}}$
$\cos \theta = \sin \theta \cdot \cot \theta = \cot \frac{\theta}{\pm \sqrt{{\cot}^{2} \theta + 1}}$
$\sec \theta = \frac{1}{\cos} \theta = \frac{\pm \sqrt{{\cot}^{2} \theta + 1}}{\cot} \theta$

Compute the values of the other remaining trigonometric functions

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

$\csc \theta = \pm \sqrt{{\cot}^{2} \theta + 1} = \pm \sqrt{{\left(\frac{- \sqrt{11}}{- 5}\right)}^{2} + 1} = \pm \sqrt{\frac{11}{25} + 1} = - \frac{6}{5}$

$\sin \theta = \frac{1}{\csc} \theta = \frac{1}{\pm \sqrt{{\cot}^{2} \theta + 1}} = \frac{1}{\pm \sqrt{{\left(\frac{- \sqrt{11}}{- 5}\right)}^{2} + 1}} = \frac{1}{\pm \sqrt{\left(\frac{11}{25} + 1\right)}} = - \frac{5}{6}$

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

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

God bless....I hope the explanation si useful.