Answer
$\csc\theta$ = - $\frac{25}{7}$
Work Step by Step
We know from third Pythagorean identity that-
$\csc\theta$ = ± $\sqrt (1+\cot^{2}\theta)$
As $\theta$ terminates in Q IV, Therefore $\csc\theta$ will be negative-
$\csc\theta$ = - $\sqrt (1+\cot^{2}\theta)$
substitute the given value of $\cot\theta$-
$\csc\theta$ = - $\sqrt (1+(\frac{-24}{7})^{2})$
$\csc\theta$= - $\sqrt (1+\frac{576}{49})$
$\csc\theta$ = - $\sqrt (\frac{49+576}{49})$ = $\sqrt (\frac{625}{49})$
$\csc\theta$ = - $\frac{25}{7}$
