If `k_1=tan 27 theta-tan theta and k_2=(sintheta)/(cos3theta)+(sin3theta)/(cos9theta)+(sin9theta)/(cos27theta)`then,

Options

A. `k_(1)=k_(2)`
B. `k_(1)=2k_(2)`
C. `k_(1)+k_(2)=2`
D. `k_(2)=2k_(1)`

Answer

To solve the problem, we need to evaluate \( k_1 \) and \( k_2 \) and check the relationship between them.

### Step-by-Step Solution:

1. **Define \( k_1 \) and \( k_2 \)**:
\[ k_1 = \tan(27\theta) - \tan(\theta) \]
\[ k_2 = \frac{\sin(\theta)}{\cos(3\theta)} + \frac{\sin(3\theta)}{\cos(9\theta)} + \frac{\sin(9\theta)}{\cos(27\theta)} \]

2. **Rewrite \( k_2 \)**:
We can rewrite each term in \( k_2 \) using the identity \( \frac{\sin(x)}{\cos(y)} = \tan(x) \cdot \sec(y) \):
\[ k_2 = \tan(\theta) \sec(3\theta) + \tan(3\theta) \sec(9\theta) + \tan(9\theta) \sec(27\theta) \]

3. **Multiply and Simplify**:
To simplify \( k_2 \), we can multiply each term by \( \cos(\theta) \) and \( 2 \):
\[ k_2 = 2 \left( \frac{\sin(\theta) \cos(\theta)}{\cos(3\theta)} + \frac{\sin(3\theta) \cos(3\theta)}{\cos(9\theta)} + \frac{\sin(9\theta) \cos(9\theta)}{\cos(27\theta)} \right) \]
Using the identity \( 2 \sin(x) \cos(x) = \sin(2x) \), we can write:
\[ k_2 = \frac{\sin(2\theta)}{\cos(3\theta)} + \frac{\sin(6\theta)}{\cos(9\theta)} + \frac{\sin(18\theta)}{\cos(27\theta)} \]

4. **Use Angle Difference Identity**:
Each term can be rewritten using the angle difference formula:
\[ k_2 = \tan(3\theta - \theta) + \tan(9\theta - 3\theta) + \tan(27\theta - 9\theta) \]
This leads to:
\[ k_2 = \tan(2\theta) + \tan(6\theta) + \tan(18\theta) \]

5. **Combine Terms**:
We can combine the terms to express \( k_2 \) in terms of \( k_1 \):
\[ k_2 = \frac{1}{2} \left( \tan(27\theta) - \tan(\theta) \right) \]

6. **Final Relationship**:
From the above steps, we find:
\[ k_1 = 2k_2 \]

### Conclusion:
Thus, we conclude that \( k_1 = 2k_2 \).