Answer
$\sin\theta$ = $- \frac{4}{5}$
$\cos\theta$ =$ -\frac{3}{5}$
$\tan\theta$ =$\frac{4}{3}$ (given)
$\cot\theta$ =$ \frac{3}{4}$
$\sec\theta$ =$ -\frac{5}{3}$
$\csc\theta$ =$- \frac{5}{4}$
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 $\tan\theta$ =$ \frac{y}{x}$ = $ \frac{4}{3}$ (Ignoring sign)
Using Pythagorous theorem-
$r =\sqrt {x^{2}+y^{2}}$ =$ \sqrt {3^{2}+4^{2}}$ = 5
i.e. x=3, y=4 and r=5
As $\theta$ lies in quadrant III, therefore-
$\sin\theta$ =$ -\frac{y}{r}$ = $- \frac{4}{5}$
$\cos\theta$ =$- \frac{x}{r}$ =$ -\frac{3}{5}$
$\tan\theta$ =$ \frac{y}{x}$ =$\frac{4}{3}$ (given)
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{3}{4}$
$\sec\theta$ =$ -\frac{r}{x}$ =$ -\frac{5}{3}$
$\csc\theta$ =$ -\frac{r}{y}$ =$- \frac{5}{4}$
