Answer
$\csc\theta$ = $\frac{29}{20}$
Work Step by Step
We know from third Pythagorean identity that-
$\csc\theta$ = ± $\sqrt (1+\cot^{2}\theta)$
As $\sin\theta\gt0$, i.e. $\sin\theta$ is positive and $\cot\theta$ is negative, hence $\theta$ terminates in Q II, Therefore $\csc\theta$ will be positive-
$\csc\theta$ = $\sqrt (1+\cot^{2}\theta)$
substitute the given value of $\cot\theta$-
$\csc\theta$ = $\sqrt (1+(\frac{-21}{20})^{2})$
$\csc\theta$= $\sqrt (1+\frac{441}{400})$
$\csc\theta$ = $\sqrt (\frac{400+441}{400})$ = $\sqrt (\frac{841}{400})$
$\csc\theta$ = $\frac{29}{20}$
