write the first expression in terms of the second if the terminal point determined by t is in the given…

Explanation:

Step1: Recall trigonometric identities

We know that $\sec(t)=\frac{1}{\cos(t)}$, so $\cos(t)=\frac{1}{\sec(t)}$, and also $\sin^{2}(t)+\cos^{2}(t) = 1$.

Step2: Express $\sin^{2}(t)$ in terms of $\cos(t)$

From $\sin^{2}(t)+\cos^{2}(t)=1$, we can get $\sin^{2}(t)=1 - \cos^{2}(t)$.

Step3: Substitute $\cos(t)$ with $\frac{1}{\sec(t)}$

Substitute $\cos(t)=\frac{1}{\sec(t)}$ into $\sin^{2}(t)=1 - \cos^{2}(t)$, we have $\sin^{2}(t)=1-\frac{1}{\sec^{2}(t)}=\frac{\sec^{2}(t)- 1}{\sec^{2}(t)}$.

Step4: Determine the sign of $\sin(t)$ in Quadrant IV

In Quadrant IV, $\sin(t)<0$. So $\sin(t)=-\sqrt{\frac{\sec^{2}(t)-1}{\sec^{2}(t)}}=-\frac{\sqrt{\sec^{2}(t)-1}}{\vert\sec(t)\vert}$. Since $\sec(t)>0$ in Quadrant IV, $\sin(t)=-\frac{\sqrt{\sec^{2}(t)-1}}{\sec(t)}$.

Answer:

$-\frac{\sqrt{\sec^{2}(t)-1}}{\sec(t)}$