`(" sin 8x cos x- sin 6x cos 3x")/(" cos 2x cos x - sin 4x sin 3x")= `

Options

A. ` " sin " 2x`
B. ` " cos " 2x`
C. ` " tan " 2x`
D. ` " cosec " 2x`

Explanation

To solve the given problem step by step, we will simplify the expression step by step using trigonometric identities.

**Given Expression:**
\[ \frac{\sin 8x \cos x - \sin 6x \cos 3x}{\cos 2x \cos x - \sin 4x \sin 3x} \]

### Step 1: Simplify the Numerator
We can use the identity for the difference of products of sine and cosine:
\[ \sin A \cos B - \sin C \cos D = \sin A \cos B - \sin C \cos D \]
We will multiply and divide the numerator by 2:
\[ = \frac{2(\sin 8x \cos x - \sin 6x \cos 3x)}{2} \]

Using the identity \(2 \sin A \cos B = \sin(A + B) + \sin(A - B)\):
\[ = \frac{\sin(8x + x) - \sin(8x - x) - \sin(6x + 3x) + \sin(6x - 3x)}{2} \]
This simplifies to:
\[ = \frac{\sin 9x - \sin 7x - \sin 9x + \sin 3x}{2} \]
Thus, the numerator simplifies to:
\[ = \frac{\sin 7x - \sin 3x}{2} \]

### Step 2: Simplify the Denominator
Now, we simplify the denominator:
\[ \cos 2x \cos x - \sin 4x \sin 3x \]
Again, we will multiply and divide the denominator by 2:
\[ = \frac{2(\cos 2x \cos x - \sin 4x \sin 3x)}{2} \]

Using the identity \(2 \cos A \cos B = \cos(A + B) + \cos(A - B)\):
\[ = \frac{\cos(2x + x) + \cos(2x - x) - (\cos(4x + 3x) + \cos(4x - 3x))}{2} \]
This simplifies to:
\[ = \frac{\cos 3x + \cos x - \cos 7x - \cos x}{2} \]
Thus, the denominator simplifies to:
\[ = \frac{\cos 3x - \cos 7x}{2} \]

### Step 3: Combine the Results
Now we can combine the simplified numerator and denominator:
\[ = \frac{\frac{\sin 7x - \sin 3x}{2}}{\frac{\cos 3x - \cos 7x}{2}} = \frac{\sin 7x - \sin 3x}{\cos 3x - \cos 7x} \]

### Step 4: Use Sine and Cosine Difference Identities
Using the identities for sine and cosine differences:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]

Applying these identities:
\[ = \frac{2 \cos\left(\frac{7x + 3x}{2}\right) \sin\left(\frac{7x - 3x}{2}\right)}{-2 \sin\left(\frac{7x + 3x}{2}\right) \sin\left(\frac{7x - 3x}{2}\right)} \]
This simplifies to:
\[ = -\frac{\cos(5x)}{\sin(2x)} = -\cot(2x) \]

### Final Result
Thus, the final expression simplifies to:
\[ \tan(2x) \]

### Conclusion
Hence, we have shown that:
\[ \frac{\sin 8x \cos x - \sin 6x \cos 3x}{\cos 2x \cos x - \sin 4x \sin 3x} = \tan(2x) \]