If for $$x \in \left(0, \frac{\pi}{2}\right)$$, $$\log_{10} \sin x + \log_{10} \cos x = -1$$ and $$\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$$, $$n > 0$$, then the value of $$n$$ is equal to:

We are given that $$\log_{10} \sin x + \log_{10} \cos x = -1$$ for $$x \in \left(0, \frac{\pi}{2}\right)$$. This means $$\log_{10}(\sin x \cos x) = -1$$, so $$\sin x \cos x = 10^{-1} = \frac{1}{10}$$.

We know that $$\sin x \cos x = \frac{\sin 2x}{2}$$, so $$\frac{\sin 2x}{2} = \frac{1}{10}$$, giving $$\sin 2x = \frac{1}{5}$$.

Now we need to find $$(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2 \times \frac{1}{10} = 1 + \frac{1}{5} = \frac{6}{5}$$.

Since $$x \in \left(0, \frac{\pi}{2}\right)$$, both $$\sin x$$ and $$\cos x$$ are positive, so $$\sin x + \cos x = \sqrt{\frac{6}{5}}$$.

Now we use the second equation: $$\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$$.

The left side is $$\log_{10}\sqrt{\frac{6}{5}} = \frac{1}{2}\log_{10}\frac{6}{5}$$.

So $$\frac{1}{2}\log_{10}\frac{6}{5} = \frac{1}{2}(\log_{10} n - 1)$$.

Multiplying both sides by 2: $$\log_{10}\frac{6}{5} = \log_{10} n - 1$$.

This gives $$\log_{10} n = 1 + \log_{10}\frac{6}{5} = \log_{10} 10 + \log_{10}\frac{6}{5} = \log_{10}\left(10 \times \frac{6}{5}\right) = \log_{10} 12$$.

Therefore $$n = 12$$, and the correct answer is option (2).