`tan5xtan3xtan2x=?`

Options

A. `tan5x-tan3x-tan2x`
B. 0
C. `(sin6x-sin3x-sin2x)/(cos5x-cos3x-cos2x)`
D. `tan9x`

Answer

To solve the equation \( \tan 5x \tan 3x \tan 2x \), we can follow these steps:

### Step 1: Rewrite \( \tan 2x \)
We can express \( \tan 2x \) in terms of \( \tan 5x \) and \( \tan 3x \). Using the tangent subtraction formula:

\[ \tan 2x = \tan(5x - 3x) = \frac{\tan 5x - \tan 3x}{1 + \tan 5x \tan 3x} \]

### Step 2: Substitute \( \tan 2x \) into the equation
Now we substitute this expression for \( \tan 2x \) into the original equation:

\[ \tan 5x \tan 3x \tan 2x = \tan 5x \tan 3x \left( \frac{\tan 5x - \tan 3x}{1 + \tan 5x \tan 3x} \right) \]

### Step 3: Simplify the equation
Now we can simplify the expression:

\[ = \frac{\tan 5x \tan 3x (\tan 5x - \tan 3x)}{1 + \tan 5x \tan 3x} \]

### Step 4: Factor the numerator
The numerator can be factored as follows:

\[ = \frac{\tan^2 5x \tan 3x - \tan 5x \tan^2 3x}{1 + \tan 5x \tan 3x} \]

### Step 5: Final expression
Thus, we have:

\[ \tan 5x \tan 3x \tan 2x = \frac{\tan^2 5x \tan 3x - \tan 5x \tan^2 3x}{1 + \tan 5x \tan 3x} \]

This is the simplified form of the expression \( \tan 5x \tan 3x \tan 2x \).