The problem requires expressing $\sin(t)$ in terms of $\sec(t)$ given that the angle $t$ is in the fourth quadrant.

Step-by-Step Solution:

1. Recall Trigonometric Identities: We know the identity: $\sin^2(t) + \cos^2(t) = 1$ Since $\sec(t) = \frac{1}{\cos(t)}$, we can express $\cos(t)$ in terms of $\sec(t)$: $\cos(t) = \frac{1}{\sec(t)}$

2. Rewrite $\sin^2(t)$: Substitute $\cos(t)$ in the identity: $\sin^2(t) = 1 - \cos^2(t) = 1 - \left(\frac{1}{\sec(t)}\right)^2$ Simplify further: $\sin^2(t) = 1 - \frac{1}{\sec^2(t)} = \frac{\sec^2(t) - 1}{\sec^2(t)}$

3. Take the Square Root: To find $\sin(t)$, take the square root of both sides: $\sin(t) = \pm \sqrt{\frac{\sec^2(t) - 1}{\sec^2(t)}}$ Simplify by expressing in terms of $\sec(t)$: $\sin(t) = \pm \frac{\sqrt{\sec^2(t) - 1}}{\sec(t)}$

4. Determine the Sign in the Fourth Quadrant: In the fourth quadrant, $\sin(t)$ is negative. Therefore: $\sin(t) = -\frac{\sqrt{\sec^2(t) - 1}}{\sec(t)}$

Final Answer:

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

Would you like a more detailed explanation, or do you have any questions?

Further Questions to Explore

1. How would you express $\cos(t)$ in terms of $\csc(t)$ in a different quadrant?
2. What happens if $t$ is in the first quadrant? How does the expression change?
3. Can you derive a similar expression for $\tan(t)$ in terms of $\sec(t)$?
4. Why does the sign of $\sin(t)$ change in different quadrants?
5. How does knowing the quadrant simplify solving trigonometric expressions?

Tip

When working with trigonometric identities in specific quadrants, always check the sign of the function based on the quadrant to avoid mistakes.