The principal value of $$\tan^{-1}\left(\cot\frac{43\pi}{4}\right)$$ is:

We need to find the principal value of $$\tan^{-1}\left(\cot\frac{43\pi}{4}\right)$$.

First, recall that $$\cot\theta = \tan\left(\frac{\pi}{2} - \theta\right)$$. So, we can write:

$$\cot\frac{43\pi}{4} = \tan\left(\frac{\pi}{2} - \frac{43\pi}{4}\right)$$

Now, compute the expression inside the tangent:

$$\frac{\pi}{2} - \frac{43\pi}{4} = \frac{2\pi}{4} - \frac{43\pi}{4} = \frac{2\pi - 43\pi}{4} = \frac{-41\pi}{4}$$

So, $$\cot\frac{43\pi}{4} = \tan\left(-\frac{41\pi}{4}\right)$$.

Since tangent is an odd function, $$\tan(-\theta) = -\tan\theta$$, so:

$$\tan\left(-\frac{41\pi}{4}\right) = -\tan\left(\frac{41\pi}{4}\right)$$

Thus, $$\cot\frac{43\pi}{4} = -\tan\left(\frac{41\pi}{4}\right)$$.

Next, simplify $$\tan\left(\frac{41\pi}{4}\right)$$. The tangent function has a period of $$\pi$$, so we reduce the angle modulo $$\pi$$.

Write $$\frac{41\pi}{4}$$ as:

$$\frac{41\pi}{4} = \frac{41}{4}\pi = 10\pi + \frac{\pi}{4}$$

because $$10\pi = \frac{40\pi}{4}$$, and $$\frac{41\pi}{4} - \frac{40\pi}{4} = \frac{\pi}{4}$$.

Since $$\tan(\theta + k\pi) = \tan\theta$$ for any integer $$k$$, and $$10\pi$$ is a multiple of $$\pi$$ (with $$k=10$$):

$$\tan\left(10\pi + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$$

Therefore, $$\tan\left(\frac{41\pi}{4}\right) = 1$$, and:

$$\cot\frac{43\pi}{4} = -\tan\left(\frac{41\pi}{4}\right) = -1$$

So, the expression becomes:

$$\tan^{-1}\left(\cot\frac{43\pi}{4}\right) = \tan^{-1}(-1)$$

The principal value of $$\tan^{-1}(x)$$ lies in the interval $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$. We know that $$\tan\left(-\frac{\pi}{4}\right) = -1$$, and $$-\frac{\pi}{4}$$ is within this interval. Hence, the principal value is $$-\frac{\pi}{4}$$.

Now, comparing with the options:

A. $$\frac{\pi}{4}$$

B. $$-\frac{\pi}{4}$$

C. $$\frac{3\pi}{4}$$

D. $$-\frac{3\pi}{4}$$

Hence, the correct answer is Option B.