Answer
$\sin\theta$ = $ \frac{1}{2}$
$\cos\theta$ =$ \frac{\sqrt 3}{2}$
$\tan\theta$ =$ \frac{1}{\sqrt 3}$
$\cot\theta$ = $\sqrt 3$
$\sec\theta$ =$ \frac{2}{\sqrt 3}$
$\csc\theta$ = 2 (given)
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 $\csc\theta$ =$ \frac{r}{y}$ = $ \frac{2}{1}$ (Ignoring sign)
Using Pythagorous theorem-
$x =\sqrt {r^{2}-y^{2}}$ =$ \sqrt {2^{2}-1^{2}}$ = $\sqrt 3$
i.e. x= $\sqrt 3$, y= 1 and r=2
Given $\theta$ lies in quadrant I, hence values of all t-functions will be positive-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{1}{2}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{\sqrt 3}{2}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{1}{\sqrt 3}$
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{\sqrt 3}{1}$ = $\sqrt 3$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{2}{\sqrt 3}$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{2}{1}$ = 2 (given)
