Answer
$\sin{\theta} = \frac{8}{10} = \frac{4}{5}
\\\cos{\theta} = \frac{-6}{10} = -\frac{3}{5}
\\\tan{\theta} = \frac{8}{-6} = -\frac{4}{3}
\\\cot{\theta}=\frac{-6}{8} = -\frac{3}{4}
\\\csc{\theta}=\frac{10}{8}=\frac{5}{4}
\\\sec{\theta} = \frac{10}{-6} - -\frac{5}{3}$
Work Step by Step
RECALL:
If P(x, y) is on the terminal side of an angle $\theta$ in standard position then:
$\begin{array}{cc}
&\sin{\theta} = \frac{y}{r} &\csc{\theta} = \frac{r}{y}
\\&\cos{\theta} = \frac{x}{r} &\sec{\theta} = \frac{r}{x}
\\&\tan{\theta} = \frac{y}{x} &\cot{\theta} = \frac{x}{y}
\end{array}$
where
$r= \sqrt{x^2+y^2}$
Solve for $r$ to obtain:
$r=\sqrt{(-6)^2+8^2} = \sqrt{36+64}=\sqrt{100}=10$
Thus, using the ratios in the recall part above, and with P(-6, 8) and $r=10$, gives:
$\sin{\theta} = \frac{8}{10} = \frac{4}{5}
\\\cos{\theta} = \frac{-6}{10} = -\frac{3}{5}
\\\tan{\theta} = \frac{8}{-6} = -\frac{4}{3}
\\\cot{\theta}=\frac{-6}{8} = -\frac{3}{4}
\\\csc{\theta}=\frac{10}{8}=\frac{5}{4}
\\\sec{\theta} = \frac{10}{-6} - -\frac{5}{3}$
