Answer
$\sin\theta$ = $- \frac{9}{\sqrt {145}}$
$\cos\theta$ =$ \frac{8}{\sqrt {145}}$
$\tan\theta$ =$- \frac{9}{8}$
$\cot\theta$ = =$- \frac{8}{9}$ (given)
$\sec\theta$ =$ \frac{\sqrt {145}}{8}$
$\csc\theta$ =$- \frac{\sqrt {145}}{9}$
Work Step by Step
Considering $\theta$ as reference angle, we may draw a right triangle ABC to calculate other trigonometric functions-
In right triangle ABC-
Given $\cot\theta$ =$ \frac{x}{y}$ = $ \frac{8}{9}$ (Ignoring sign)
Using Pythagorous theorem-
$r =\sqrt {x^{2}+y^{2}}$ =$ \sqrt {8^{2}+9^{2}}$ = $\sqrt {145}$
i.e. x= 8, y= 9 and r=$\sqrt {145}$
Given cot is negative and cos is positive, therefore $\theta$ lies in quadrant IV, hence-
$\sin\theta$ =$ -\frac{y}{r}$ = $- \frac{9}{\sqrt {145}}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{8}{\sqrt {145}}$
$\tan\theta$ =$ -\frac{y}{x}$ =$- \frac{9}{8}$
$\cot\theta$ =$ -\frac{x}{y}$ =$- \frac{8}{9}$ (given)
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{\sqrt {145}}{8}$
$\csc\theta$ =$ -\frac{r}{y}$ =$- \frac{\sqrt {145}}{9}$