The function f(x) is defined as f(x) = (tan x + tan x) / sin x for x ≠ 0, and f(0) = a. Find the value of 'a' that makes the function continuous at x=0.

For a function $f(x)$ to be continuous at $x=0$, the limit of the function as $x$ approaches 0 must be equal to the function's value at $x=0$.
Given $f(0) = a$, we must have:

$$a = \lim_{x \to 0} f(x)$$

The given function is $f(x) = \frac{\text{ظا س} + \text{ظا س}}{\text{جا س}}$ for $x \neq 0$.
This can be written as:

$$f(x) = \frac{2 \tan x}{\sin x}$$

Substitute the simplified function into the limit expression:

$$a = \lim_{x \to 0} \frac{2 \tan x}{\sin x}$$

Recall that $\tan x = \frac{\sin x}{\cos x}$. Substitute this identity into the expression:

$$a = \lim_{x \to 0} \frac{2 \left(\frac{\sin x}{\cos x}\right)}{\sin x}$$

Since $x \neq 0$, $\sin x \neq 0$, so we can cancel $\sin x$ from the numerator and denominator:

$$a = \lim_{x \to 0} \frac{2}{\cos x}$$

$$a = \frac{2}{\cos(0)}$$

We know that $\cos(0) = 1$.

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

The value of $a$ is 2.
Comparing this with the given options, option 2 is 2.

The final answer is $\text{2}$.