`(sin 3theta+sin5theta+sin7theta+sin9theta)/(cos 3theta+cos 5theta+cos 7theta+cos9theta)` is equal to

Options

A. `tan 3theta`
B. `cot 3theta`
C. `tan 6 theta`
D. `cot 6 theta`

Explanation

To solve the expression \(\frac{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta}{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta}\), we will follow these steps:

### Step 1: Grouping the Sine and Cosine Terms
We can group the sine and cosine terms as follows:
\[ \frac{(\sin 3\theta + \sin 9\theta) + (\sin 5\theta + \sin 7\theta)}{(\cos 3\theta + \cos 9\theta) + (\cos 5\theta + \cos 7\theta)} \]

### Step 2: Applying the Sine Addition Formula
Using the sine addition formula:
\[ \sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]
We apply this to \(\sin 3\theta + \sin 9\theta\):
- Here, \(a = 3\theta\) and \(b = 9\theta\).
\[ \sin 3\theta + \sin 9\theta = 2 \sin\left(\frac{3\theta + 9\theta}{2}\right) \cos\left(\frac{3\theta - 9\theta}{2}\right) = 2 \sin(6\theta) \cos(-3\theta) = 2 \sin(6\theta) \cos(3\theta) \]

Now applying the same formula to \(\sin 5\theta + \sin 7\theta\):
- Here, \(a = 5\theta\) and \(b = 7\theta\).
\[ \sin 5\theta + \sin 7\theta = 2 \sin\left(\frac{5\theta + 7\theta}{2}\right) \cos\left(\frac{5\theta - 7\theta}{2}\right) = 2 \sin(6\theta) \cos(-\theta) = 2 \sin(6\theta) \cos(\theta) \]

### Step 3: Combining the Sine Terms
Now, substituting back into the numerator:
\[ \sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta = 2 \sin(6\theta) \cos(3\theta) + 2 \sin(6\theta) \cos(\theta) = 2 \sin(6\theta) (\cos(3\theta) + \cos(\theta)) \]

### Step 4: Applying the Cosine Addition Formula
Using the cosine addition formula:
\[ \cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]
We apply this to \(\cos 3\theta + \cos 9\theta\):
\[ \cos 3\theta + \cos 9\theta = 2 \cos(6\theta) \cos(-3\theta) = 2 \cos(6\theta) \cos(3\theta) \]

Now applying the same formula to \(\cos 5\theta + \cos 7\theta\):
\[ \cos 5\theta + \cos 7\theta = 2 \cos(6\theta) \cos(-\theta) = 2 \cos(6\theta) \cos(\theta) \]

### Step 5: Combining the Cosine Terms
Now, substituting back into the denominator:
\[ \cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta = 2 \cos(6\theta) \cos(3\theta) + 2 \cos(6\theta) \cos(\theta) = 2 \cos(6\theta) (\cos(3\theta) + \cos(\theta)) \]

### Step 6: Simplifying the Expression
Now we can rewrite the entire expression:
\[ \frac{2 \sin(6\theta) (\cos(3\theta) + \cos(\theta))}{2 \cos(6\theta) (\cos(3\theta) + \cos(\theta))} \]
We can cancel \(2\) and \((\cos(3\theta) + \cos(\theta))\) from the numerator and denominator (assuming \(\cos(3\theta) + \cos(\theta) \neq 0\)):
\[ = \frac{\sin(6\theta)}{\cos(6\theta)} = \tan(6\theta) \]

### Final Answer
Thus, the expression simplifies to:
\[ \tan(6\theta) \]