If $$\tan 15° + \frac{1}{\tan 75°} + \frac{1}{\tan 105°} + \tan 195° = 2a$$, then the value of $$a + \frac{1}{a}$$ is:

We need to find $$a + \frac{1}{a}$$ given that $$\tan 15° + \frac{1}{\tan 75°} + \frac{1}{\tan 105°} + \tan 195° = 2a$$.

To begin,

$$\frac{1}{\tan 75°} = \frac{1}{\cot 15°} = \tan 15°$$

$$\frac{1}{\tan 105°} = \frac{1}{\tan(180° - 75°)} = \frac{1}{-\tan 75°} = -\frac{1}{\tan 75°} = -\tan 15°$$

$$\tan 195° = \tan(180° + 15°) = \tan 15°$$

Next,

$$\tan 15° + \tan 15° - \tan 15° + \tan 15° = 2\tan 15°$$

So $$2a = 2\tan 15°$$, giving $$a = \tan 15°$$.

From this,

$$\tan 15° = 2 - \sqrt{3}$$

$$\frac{1}{\tan 15°} = \frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}$$

$$a + \frac{1}{a} = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$$

The correct answer is Option 1: $$4$$.