`(sin3theta+sin5theta+sin7theta+sin9theta)/(cos3theta+cos5theta+cos7theta+cos9theta)=?`

Options

A. `cot6theta`
B. `tan6theta`
C. `cot3theta`
D. `tan3theta`

Answer

To solve the problem \((\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta) / (\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta)\), we will use trigonometric identities to simplify the expression step by step.

### Step-by-Step Solution:

1. **Group the Sine Terms:**
We can group the sine terms into pairs:
\[ \sin 3\theta + \sin 9\theta \quad \text{and} \quad \sin 5\theta + \sin 7\theta \]

2. **Apply the Sine Addition Formula:**
Using the formula \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\):
- For \(\sin 3\theta + \sin 9\theta\):
\[ \sin 3\theta + \sin 9\theta = 2 \sin\left(\frac{3\theta + 9\theta}{2}\right) \cos\left(\frac{9\theta - 3\theta}{2}\right) = 2 \sin(6\theta) \cos(3\theta) \]
- For \(\sin 5\theta + \sin 7\theta\):
\[ \sin 5\theta + \sin 7\theta = 2 \sin\left(\frac{5\theta + 7\theta}{2}\right) \cos\left(\frac{7\theta - 5\theta}{2}\right) = 2 \sin(6\theta) \cos(\theta) \]

3. **Combine the Results:**
Now, we can combine these results in the numerator:
\[ \sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta = 2 \sin(6\theta) \cos(3\theta) + 2 \sin(6\theta) \cos(\theta) \]
Factoring out \(2 \sin(6\theta)\):
\[ = 2 \sin(6\theta) (\cos(3\theta) + \cos(\theta)) \]

4. **Group the Cosine Terms:**
Similarly, group the cosine terms:
\[ \cos 3\theta + \cos 9\theta \quad \text{and} \quad \cos 5\theta + \cos 7\theta \]

5. **Apply the Cosine Addition Formula:**
Using the formula \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\):
- For \(\cos 3\theta + \cos 9\theta\):
\[ \cos 3\theta + \cos 9\theta = 2 \cos(6\theta) \cos(3\theta) \]
- For \(\cos 5\theta + \cos 7\theta\):
\[ \cos 5\theta + \cos 7\theta = 2 \cos(6\theta) \cos(\theta) \]

6. **Combine the Results:**
Now, we can combine these results in the denominator:
\[ \cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta = 2 \cos(6\theta) \cos(3\theta) + 2 \cos(6\theta) \cos(\theta) \]
Factoring out \(2 \cos(6\theta)\):
\[ = 2 \cos(6\theta) (\cos(3\theta) + \cos(\theta)) \]

7. **Final Expression:**
Now we can rewrite the original expression:
\[ \frac{2 \sin(6\theta) (\cos(3\theta) + \cos(\theta))}{2 \cos(6\theta) (\cos(3\theta) + \cos(\theta))} \]
The \(2\) and \((\cos(3\theta) + \cos(\theta))\) terms cancel out:
\[ = \frac{\sin(6\theta)}{\cos(6\theta)} = \tan(6\theta) \]

### Final Answer:
\[ \frac{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta}{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta} = \tan(6\theta) \]