Let's solve each of these trigonometric function values for $-\frac{3\pi}{4}$.

(a) $\tan\left(-\frac{3\pi}{4}\right)$

The tangent function is odd, meaning $\tan(-x) = -\tan(x)$. Therefore: $\tan\left(-\frac{3\pi}{4}\right) = -\tan\left(\frac{3\pi}{4}\right)$ Now, $\frac{3\pi}{4}$ is in the second quadrant, where the tangent function is negative. The reference angle is $\frac{\pi}{4}$, so: $\tan\left(\frac{3\pi}{4}\right) = -1$ Thus: $\tan\left(-\frac{3\pi}{4}\right) = -(-1) = 1$

(b) $\csc\left(-\frac{3\pi}{4}\right)$

The cosecant function is odd as well, meaning $\csc(-x) = -\csc(x)$. Therefore: $\csc\left(-\frac{3\pi}{4}\right) = -\csc\left(\frac{3\pi}{4}\right)$ In the second quadrant, where $\frac{3\pi}{4}$ lies, the cosecant function is positive. Using the reference angle $\frac{\pi}{4}$, we know: $\csc\left(\frac{3\pi}{4}\right) = \csc\left(\frac{\pi}{4}\right) = \sqrt{2}$ Thus: $\csc\left(-\frac{3\pi}{4}\right) = -\sqrt{2}$

(c) $\cot\left(-\frac{3\pi}{4}\right)$

The cotangent function is odd, so $\cot(-x) = -\cot(x)$. Thus: $\cot\left(-\frac{3\pi}{4}\right) = -\cot\left(\frac{3\pi}{4}\right)$ Since $\cot\left(\frac{3\pi}{4}\right) = -1$ (as it’s the reciprocal of $\tan\left(\frac{3\pi}{4}\right)$), we get: $\cot\left(-\frac{3\pi}{4}\right) = -(-1) = 1$

Summary of Answers:

1. $\tan\left(-\frac{3\pi}{4}\right) = 1$
2. $\csc\left(-\frac{3\pi}{4}\right) = -\sqrt{2}$
3. $\cot\left(-\frac{3\pi}{4}\right) = 1$

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Related Questions:

1. How would these values change if the angle were positive?
2. What is the significance of odd and even functions in trigonometry?
3. How do we determine reference angles for non-standard angles?
4. What are the values of other trigonometric functions at $-\frac{3\pi}{4}$?
5. How would we solve this problem in terms of degrees?

Tip:

For angles in the unit circle, knowing the quadrant helps determine the sign of each trigonometric function.