1. Simplify each trigonometric term

We use standard identities:

• $\tan(180^\circ+\theta)=\tan\theta$
• $\tan(90^\circ-\theta)=\cot\theta=\dfrac{1}{\tan\theta}$
• $\tan(180^\circ-\theta)=-\tan\theta$

Given

$$\tan 15^\circ + \frac{1}{\tan 75^\circ} + \frac{1}{\tan 105^\circ} + \tan 195^\circ = 2a$$

Now simplify term by term:

• $\dfrac{1}{\tan 75^\circ}=\cot 75^\circ=\tan 15^\circ$

• $\tan 105^\circ=\tan(180^\circ-75^\circ)=-\tan 75^\circ$

  so,

  $$\frac{1}{\tan 105^\circ}=-\frac{1}{\tan 75^\circ}=-\tan 15^\circ$$

• $\tan 195^\circ=\tan(180^\circ+15^\circ)=\tan 15^\circ$

So the left-hand side becomes

$$\tan 15^\circ + \tan 15^\circ - \tan 15^\circ + \tan 15^\circ = 2\tan 15^\circ$$

Hence,

$$2a=2\tan 15^\circ \implies a=\tan 15^\circ$$

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2. Find $\tan 15^\circ$

Using

$$\tan(45^\circ-30^\circ)=\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ}$$

we get

$$\tan15^\circ=\frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}} =\frac{\sqrt3-1}{\sqrt3+1}$$

Rationalizing,

$$\tan15^\circ=\frac{(\sqrt3-1)^2}{3-1} =\frac{3+1-2\sqrt3}{2} =2-\sqrt3$$

Thus,

$$a=2-\sqrt3$$

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3. Compute $a+\dfrac{1}{a}$

$$\frac{1}{a}=\frac{1}{2-\sqrt3}$$

Rationalizing,

$$\frac{1}{2-\sqrt3}=\frac{2+\sqrt3}{(2-\sqrt3)(2+\sqrt3)}=2+\sqrt3$$

Therefore,

$$a+\frac{1}{a}=(2-\sqrt3)+(2+\sqrt3)=4$$

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4. Final answer

$$\boxed{4}$$

So the correct option is D.