If `sum_(r=1)^(n)((tan 2^(r-1))/(cos2^(r )))=tanp^(n)-tan q`, then find the value of `(p+q)`.

Answer

To solve the problem, we need to evaluate the summation and compare it with the given expression. Let's break it down step by step.

### Step 1: Write the summation
We start with the given expression:
\[ \sum_{r=1}^{n} \frac{\tan(2^{r-1})}{\cos(2^{r})} \]

### Step 2: Rewrite the tangent function
Recall that \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\). Thus, we can rewrite the summation as:
\[ \sum_{r=1}^{n} \frac{\sin(2^{r-1})}{\cos(2^{r})} \]

### Step 3: Manipulate the expression
We can manipulate the expression by multiplying the numerator and denominator by \(\cos(2^{r-1})\):
\[ \sum_{r=1}^{n} \frac{\sin(2^{r-1}) \cdot \cos(2^{r-1})}{\cos(2^{r}) \cdot \cos(2^{r-1})} \]
This simplifies to:
\[ \sum_{r=1}^{n} \frac{\sin(2^{r-1})}{\cos(2^{r-1})} \cdot \frac{1}{\cos(2^{r})} \]

### Step 4: Use the sine subtraction formula
Using the identity \(\sin(a) \cos(b) - \cos(a) \sin(b) = \sin(a-b)\), we can express the summation as:
\[ \sum_{r=1}^{n} \sin(2^{r}) - \sin(2^{r-1}) \]

### Step 5: Evaluate the summation
This is a telescoping series. Most terms will cancel out:
\[ \sin(2^{n}) - \sin(2^{0}) = \sin(2^{n}) - \sin(1) \]

### Step 6: Compare with the given expression
Now we have:
\[ \sin(2^{n}) - \sin(1) = \tan(p^{n}) - \tan(q) \]
We can identify that:
\[ \tan(p^{n}) = \sin(2^{n}) \quad \text{and} \quad \tan(q) = \sin(1) \]

### Step 7: Find values of p and q
From the above, we can deduce:
- \(p = 2\) (since \(2^{n}\) corresponds to \(p^{n}\))
- \(q = 1\) (since \(1\) corresponds to \(q\))

### Step 8: Calculate \(p + q\)
Finally, we find:
\[ p + q = 2 + 1 = 3 \]

### Final Answer
Thus, the value of \(p + q\) is:
\[ \boxed{3} \]