# How do you find the other five trigonometric values given csc b = -6/5 and cot b > 0?

Jul 3, 2016

$\sin b = - \frac{5}{6}$, $\cos b = - \frac{\sqrt{11}}{6}$, $\tan b = \frac{5}{\sqrt{11}}$,

$\cot b = \frac{\sqrt{11}}{5}$ and $\sec b = - \frac{6}{\sqrt{11}}$.

#### Explanation:

As $\csc b = - \frac{6}{5}$ $\implies$ ${\csc}^{2} b = \frac{36}{25}$ and using ${\csc}^{2} b = 1 + {\cot}^{2} b$, we have

$1 + {\cot}^{2} b = \frac{36}{25}$ and $\cot b = \sqrt{\frac{36}{25} - 1} = \frac{\sqrt{11}}{5}$ (as $\cot b > 0$)

As $\cot b = \frac{\sqrt{11}}{5}$, $\tan b = \frac{1}{\cot} b = \frac{5}{\sqrt{11}}$

As $\csc b = - \frac{6}{5}$, $\sin b = \frac{1}{\csc} b = - \frac{5}{6}$

and $\cos b = \cos \frac{b}{\sin} b \times \sin b = \cot b \times \sin b$

= $\frac{\sqrt{11}}{5} \times \left(- \frac{5}{6}\right) = - \frac{\sqrt{11}}{6}$

and $\sec b = \frac{1}{\cos} b = - \frac{6}{\sqrt{11}}$.

Hence, we have $\sin b = - \frac{5}{6}$, $\cos b = - \frac{\sqrt{11}}{6}$, $\tan b = \frac{5}{\sqrt{11}}$,

$\cot b = \frac{\sqrt{11}}{5}$ and $\sec b = - \frac{6}{\sqrt{11}}$.