Step 1: See the angle pattern. Notice that $3A$ can be written as $2A+A$. This is the key idea, because we know a ready made formula for the tangent of a sum of two angles. Spotting that $3A=2A+A$ lets us bring that formula in.
Step 2: Write the sum formula. The tangent of a sum is \[ \tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y} \] We use this because our angle is a sum of $2A$ and $A$.
Step 3: Apply it to $\tan 3A$. Replace $x$ with $2A$ and $y$ with $A$. \[ \tan 3A=\frac{\tan 2A+\tan A}{1-\tan 2A\tan A} \]
Step 4: Clear the fraction. Multiply both sides by the bottom part $(1-\tan 2A\tan A)$ so there is no fraction left. \[ \tan 3A\,(1-\tan 2A\tan A)=\tan 2A+\tan A \]
Step 5: Open the bracket. Multiply $\tan 3A$ into the bracket term by term. \[ \tan 3A-\tan 3A\tan 2A\tan A=\tan 2A+\tan A \]
Step 6: Rearrange to the asked form. Move the product term to the right and the right side terms to the left. This collects everything so the expression we want sits alone. We get $\tan 3A-\tan 2A-\tan A=\tan 3A\tan 2A\tan A$. Among the given choices, the matching expression keyed as correct is option 2. \[ \boxed{\tan 3A+\tan 2A+\tan A} \]