If `P=(tan(3^(n+1)theta)-tan theta) and Q=sum_(r=0)^(n) (sin(3^(r )theta))/(cos(3^(r+1)theta))`, then

Options

A. `P=2Q`
B. `P=3Q`
C. `2P=Q`
D. `3P=Q`

Answer

To solve the problem, we need to establish a relationship between \( P \) and \( Q \) given the definitions of both.

### Given:
1. \( P = \tan(3^{(n+1)}\theta) - \tan(\theta) \)
2. \( Q = \sum_{r=0}^{n} \frac{\sin(3^r\theta)}{\cos(3^{r+1}\theta)} \)

### Step-by-Step Solution:

**Step 1: Simplifying \( Q \)**

We start by rewriting \( Q \):
\[ Q = \sum_{r=0}^{n} \frac{\sin(3^r\theta)}{\cos(3^{r+1}\theta)} \]

**Step 2: Using the identity for sine and cosine**

Using the identity \( \frac{\sin A}{\cos B} = \tan A \) when \( A = 3^r\theta \) and \( B = 3^{r+1}\theta \), we can express \( Q \) as:
\[ Q = \sum_{r=0}^{n} \tan(3^r\theta) \cdot \frac{1}{\cos(3^{r+1}\theta)} \]

**Step 3: Factor out common terms**

We can factor out \( \frac{1}{2} \) from the sum:
\[ Q = \frac{1}{2} \sum_{r=0}^{n} \left( \tan(3^{r+1}\theta) - \tan(3^r\theta) \right) \]

**Step 4: Telescoping series**

This is a telescoping series, where most terms will cancel out:
\[ Q = \frac{1}{2} \left( \tan(3^{n+1}\theta) - \tan(\theta) \right) \]

**Step 5: Relating \( Q \) to \( P \)**

Now, substituting back into the expression for \( P \):
\[ P = \tan(3^{(n+1)}\theta) - \tan(\theta) \]
Thus,
\[ Q = \frac{1}{2} P \]

**Step 6: Final relationship**

From the above, we can express \( P \) in terms of \( Q \):
\[ P = 2Q \]

### Conclusion

The relationship between \( P \) and \( Q \) is:
\[ P = 2Q \]