To solve this problem, we need to express $\sin(t)$ in terms of $\sec(t)$ given that the terminal point determined by $t$ is in Quadrant IV.

Step-by-Step Solution

1. Use the identity: $\sin^2(t) + \cos^2(t) = 1$ We can solve for $\sin(t)$ if we know $\cos(t)$.

2. Express $\cos(t)$ in terms of $\sec(t)$: $\sec(t) = \frac{1}{\cos(t)}$ So, $\cos(t) = \frac{1}{\sec(t)}$

3. Substitute $\cos(t)$ in the Pythagorean identity: $\sin^2(t) + \left(\frac{1}{\sec(t)}\right)^2 = 1$ Simplifying, we get: $\sin^2(t) + \frac{1}{\sec^2(t)} = 1$ Rearranging terms: $\sin^2(t) = 1 - \frac{1}{\sec^2(t)}$

4. Simplify further: $\sin^2(t) = \frac{\sec^2(t) - 1}{\sec^2(t)}$

5. Take the square root: Since $t$ is in Quadrant IV, where $\sin(t)$ is negative, we take the negative root: $\sin(t) = -\sqrt{\frac{\sec^2(t) - 1}{\sec^2(t)}}$ Alternatively: $\sin(t) = -\frac{\sqrt{\sec^2(t) - 1}}{\sec(t)}$

So, the answer is: $\sin(t) = -\frac{\sqrt{\sec^2(t) - 1}}{\sec(t)}$

Would you like more details, or do you have any further questions?

Related Questions

1. How would the answer change if $t$ were in Quadrant II instead of Quadrant IV?
2. How can we express $\cos(t)$ in terms of $\sec(t)$ and use it for other trigonometric identities?
3. What is the significance of choosing the positive or negative root based on the quadrant?
4. How can we find $\tan(t)$ in terms of $\sec(t)$ given the quadrant?
5. Can you derive a similar expression for $\cos(t)$ in terms of $\csc(t)$?

Tip

When using trigonometric identities, always keep track of the quadrant to determine the correct sign for each trigonometric function.