Question #153fd

Dec 30, 2016

From the above figure

Given the coordinates of point Q (3,-9)
$O N = 3 \mathmr{and} Q N = - 9$

$O Q = \sqrt{O {N}^{2} + Q {N}^{2}}$

Now

$\sin \theta = \text{opposite"/"hypotenuse} = \frac{Q N}{O Q} = - \frac{9}{3 \sqrt{10}} = - \frac{3}{\sqrt{10}}$

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

$\cos \theta = \text{adjacent"/"hypotenuse} = \frac{O N}{O Q} = \frac{3}{3 \sqrt{10}} = \frac{1}{\sqrt{10}}$

So $\sec \theta = \frac{1}{\cos} \theta = \sqrt{10}$

$\tan \theta = \text{opposite"/"adajcent} = \frac{Q N}{O N} = - \frac{9}{3} = - 3$

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