$$\tan\left(2\tan^{-1}\dfrac{1}{5} + \sec^{-1}\dfrac{\sqrt{5}}{2} + 2\tan^{-1}\dfrac{1}{8}\right)$$ is equal to:

We need to evaluate $$\tan\left(2\tan^{-1}\frac{1}{5} + \sec^{-1}\frac{\sqrt{5}}{2} + 2\tan^{-1}\frac{1}{8}\right)$$.

Using the double angle formula $$2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}$$ (valid for $$|x| < 1$$):

$$2\tan^{-1}\frac{1}{5} = \tan^{-1}\frac{2 \cdot \frac{1}{5}}{1-\frac{1}{25}} = \tan^{-1}\frac{\frac{2}{5}}{\frac{24}{25}} = \tan^{-1}\frac{2 \times 25}{5 \times 24} = \tan^{-1}\frac{50}{120} = \tan^{-1}\frac{5}{12}$$

Let $$\alpha = \sec^{-1}\frac{\sqrt{5}}{2}$$. Then $$\sec\alpha = \frac{\sqrt{5}}{2}$$, so $$\cos\alpha = \frac{2}{\sqrt{5}}$$.

Using $$\sin^2\alpha = 1 - \cos^2\alpha = 1 - \frac{4}{5} = \frac{1}{5}$$, so $$\sin\alpha = \frac{1}{\sqrt{5}}$$.

Therefore $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1/\sqrt{5}}{2/\sqrt{5}} = \frac{1}{2}$$.

So $$\sec^{-1}\frac{\sqrt{5}}{2} = \tan^{-1}\frac{1}{2}$$.

$$2\tan^{-1}\frac{1}{8} = \tan^{-1}\frac{2 \cdot \frac{1}{8}}{1-\frac{1}{64}} = \tan^{-1}\frac{\frac{1}{4}}{\frac{63}{64}} = \tan^{-1}\frac{64}{4 \times 63} = \tan^{-1}\frac{16}{63}$$

Using $$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}$$ (when $$ab < 1$$):

Here $$ab = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24} < 1$$, so the formula applies.

$$\tan^{-1}\frac{5}{12} + \tan^{-1}\frac{1}{2} = \tan^{-1}\frac{\frac{5}{12}+\frac{1}{2}}{1-\frac{5}{24}} = \tan^{-1}\frac{\frac{5+6}{12}}{\frac{24-5}{24}} = \tan^{-1}\frac{\frac{11}{12}}{\frac{19}{24}} = \tan^{-1}\frac{11 \times 24}{12 \times 19} = \tan^{-1}\frac{22}{19}$$

Check: $$\frac{22}{19} \times \frac{16}{63} = \frac{352}{1197} < 1$$, so the addition formula applies.

Numerator: $$\frac{22}{19}+\frac{16}{63} = \frac{22 \times 63 + 16 \times 19}{19 \times 63} = \frac{1386 + 304}{1197} = \frac{1690}{1197}$$

Denominator: $$1-\frac{352}{1197} = \frac{1197-352}{1197} = \frac{845}{1197}$$

$$\tan\left(\tan^{-1}\frac{22}{19} + \tan^{-1}\frac{16}{63}\right) = \frac{1690}{845} = 2$$

Therefore, the correct answer is Option B: 2.