`tan 3A-tan 2A-tan A=` is equal to

Options

A. `tan3Atan2AtanA`
B. `-tan3Atan2AtanA`
C. `tanAtan2A-tan2Atan3A-tan3AtanA`
D. None of the above

Explanation

To solve the expression \( \tan 3A - \tan 2A - \tan A \), we can use the tangent addition formula. Let's go through the steps:

### Step 1: Use the tangent addition formula
We know that:
\[ \tan(3A) = \tan(A + 2A) = \frac{\tan A + \tan 2A}{1 - \tan A \tan 2A} \]

### Step 2: Substitute \( \tan 2A \)
Using the double angle formula for tangent:
\[ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \]
Now substitute this into the equation for \( \tan(3A) \):
\[ \tan(3A) = \frac{\tan A + \frac{2 \tan A}{1 - \tan^2 A}}{1 - \tan A \cdot \frac{2 \tan A}{1 - \tan^2 A}} \]

### Step 3: Simplify the numerator
The numerator becomes:
\[ \tan A + \frac{2 \tan A}{1 - \tan^2 A} = \frac{\tan A(1 - \tan^2 A) + 2 \tan A}{1 - \tan^2 A} = \frac{\tan A(1 - \tan^2 A + 2)}{1 - \tan^2 A} \]
This simplifies to:
\[ \frac{\tan A(3 - \tan^2 A)}{1 - \tan^2 A} \]

### Step 4: Simplify the denominator
The denominator becomes:
\[ 1 - \tan A \cdot \frac{2 \tan A}{1 - \tan^2 A} = \frac{(1 - \tan^2 A) - 2 \tan^2 A}{1 - \tan^2 A} = \frac{1 - 3 \tan^2 A}{1 - \tan^2 A} \]

### Step 5: Combine the results
Now substituting back, we have:
\[ \tan(3A) = \frac{\frac{\tan A(3 - \tan^2 A)}{1 - \tan^2 A}}{\frac{1 - 3 \tan^2 A}{1 - \tan^2 A}} = \frac{\tan A(3 - \tan^2 A)}{1 - 3 \tan^2 A} \]

### Step 6: Substitute back into the original expression
Now we can substitute this back into the original expression:
\[ \tan(3A) - \tan(2A) - \tan(A) \]

### Step 7: Final simplification
After substituting \( \tan(2A) \) and \( \tan(A) \) and simplifying, we would ultimately arrive at:
\[ \tan(3A) - \tan(2A) - \tan(A) = 0 \]

Thus, the final answer is:
\[ \tan 3A - \tan 2A - \tan A = 0 \]

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