Answer
${\frac{3}{\sqrt{10}},{{\frac{-1}{\sqrt{10}}}},{{\frac{-3}{1}}}},{{\frac{\sqrt{10}}{3}}},{{\frac{-\sqrt{10}}{1}}},{{\frac{-1}{3}}}$
Work Step by Step
given x=-1 and y=3
To solve the value of six trignometric function
first value of r
$r={\sqrt{(x^2+y^2)}}$
$r={\sqrt{(1^2+(-3)^2)}} ={\sqrt{10}}$
so $\sin\theta={\frac{y}{r}={\frac{3}{\sqrt{10}}}}$
$\cos\theta={\frac{x}{r}={\frac{-1}{\sqrt{10}}}}$
$\tan\theta={\frac{y}{x}={\frac{-3}{1}}}$
$\csc\theta={\frac{r}{y}={\frac{\sqrt{10}}{3}}}$
$\sec\theta={\frac{r}{x}={\frac{-\sqrt{10}}{1}}}$
$\cot\theta={\frac{x}{y}={\frac{-1}{3}}}$
the final answer is:${\frac{3}{\sqrt{10}},{{\frac{-1}{\sqrt{10}}}},{{\frac{-3}{1}}}},{{\frac{\sqrt{10}}{3}}},{{\frac{-\sqrt{10}}{1}}},{{\frac{-1}{3}}}$
