The problem asks to express the tangent function $\tan(\theta)$ in terms of the secant function $\sec(\theta)$, and it offers several multiple-choice answers.

Step-by-Step Solution

We know from trigonometric identities that: $\tan(\theta) = \pm \sqrt{\sec^2(\theta) - 1}$

This identity comes from the Pythagorean identity: $\tan^2(\theta) + 1 = \sec^2(\theta)$ Rearranging, we get: $\tan^2(\theta) = \sec^2(\theta) - 1$ Taking the square root on both sides gives: $\tan(\theta) = \pm \sqrt{\sec^2(\theta) - 1}$

Conclusion

The correct answer is therefore: $\boxed{\pm \sqrt{\sec^2(\theta) - 1}}$ This corresponds to option B in the choices provided.

Would you like further explanation on how the identity was derived or details about tangent and secant relationships?

Related Questions

1. What is the relationship between $\sec(\theta)$ and $\cos(\theta)$?
2. How can we express $\sin(\theta)$ in terms of $\sec(\theta)$?
3. What are other Pythagorean identities involving trigonometric functions?
4. How do we handle the $\pm$ sign in trigonometric identities?
5. What is the definition of the tangent function in terms of sine and cosine?

Tip

Remember that the square root introduces a $\pm$ symbol, so it's important to consider the quadrant where $\theta$ is located to determine the correct sign in trigonometric functions.