If $$A = \begin{bmatrix} 0 & -\tan\left(\frac{\theta}{2}\right) \\ \tan\left(\frac{\theta}{2}\right) & 0 \end{bmatrix}$$ and $$(I_2 + A)(I_2 - A)^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$$, then $$13(a^2 + b^2)$$ is equal to ______.

Let $$t = \tan\left(\frac{\theta}{2}\right)$$. Then $$A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}$$.

We have $$I_2 + A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$$ and $$I_2 - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$$.

The determinant of $$I_2 - A$$ is $$1 + t^2$$, so $$(I_2 - A)^{-1} = \frac{1}{1 + t^2}\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$$.

Now $$(I_2 + A)(I_2 - A)^{-1} = \frac{1}{1+t^2}\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = \frac{1}{1+t^2}\begin{bmatrix} 1-t^2 & -2t \\ 2t & 1-t^2 \end{bmatrix}$$.

Comparing with $$\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$$, we identify $$a = \frac{1-t^2}{1+t^2}$$ and $$b = \frac{2t}{1+t^2}$$.

These are the well-known half-angle identities: $$a = \cos\theta$$ and $$b = \sin\theta$$.

Therefore, $$a^2 + b^2 = \cos^2\theta + \sin^2\theta = 1$$, and $$13(a^2 + b^2) = 13 \times 1 = 13$$.