If `sum_(n=1)^(2013)tan(theta/(2^(n)))sec((theta)/(2^(n-1))) = tan((theta)/(2^(a)))-tan((theta)/(2^(b)))` then `(b+a)` equals

Options

A. 2014
B. 2012
C. 2013
D. 2014

Answer

To solve the given equation

\[ \sum_{n=1}^{2013} \tan\left(\frac{\theta}{2^n}\right) \sec\left(\frac{\theta}{2^{n-1}}\right) = \tan\left(\frac{\theta}{2^a}\right) - \tan\left(\frac{\theta}{2^b}\right), \]

we will follow these steps:

### Step 1: Rewrite the Left-Hand Side
We start with the left-hand side:

\[ \sum_{n=1}^{2013} \tan\left(\frac{\theta}{2^n}\right) \sec\left(\frac{\theta}{2^{n-1}}\right). \]

Recall that \(\sec(x) = \frac{1}{\cos(x)}\), so we can rewrite the term:

\[ \sec\left(\frac{\theta}{2^{n-1}}\right) = \frac{1}{\cos\left(\frac{\theta}{2^{n-1}}\right)}. \]

Thus, we have:

\[ \sum_{n=1}^{2013} \tan\left(\frac{\theta}{2^n}\right) \cdot \frac{1}{\cos\left(\frac{\theta}{2^{n-1}}\right)}. \]

### Step 2: Use the Identity for \(\tan\) and \(\sec\)
Using the identity \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), we can express the left-hand side as:

\[ \sum_{n=1}^{2013} \frac{\sin\left(\frac{\theta}{2^n}\right)}{\cos\left(\frac{\theta}{2^n}\right) \cos\left(\frac{\theta}{2^{n-1}}\right)}. \]

### Step 3: Simplify the Sum
We can simplify the sum further. Notice that:

\[ \tan\left(\frac{\theta}{2^n}\right) \sec\left(\frac{\theta}{2^{n-1}}\right) = \frac{\sin\left(\frac{\theta}{2^n}\right)}{\cos\left(\frac{\theta}{2^n}\right)} \cdot \frac{1}{\cos\left(\frac{\theta}{2^{n-1}}\right)}. \]

This can be rewritten as:

\[ \frac{\sin\left(\frac{\theta}{2^n}\right)}{\cos\left(\frac{\theta}{2^n}\right) \cos\left(\frac{\theta}{2^{n-1}}\right)} = \tan\left(\frac{\theta}{2^n}\right) \sec\left(\frac{\theta}{2^{n-1}}\right). \]

### Step 4: Recognize the Pattern
Notice that this sum telescopes. Each term cancels with the next, leading to:

\[ \tan\left(\frac{\theta}{2^1}\right) - \tan\left(\frac{\theta}{2^{2014}}\right). \]

### Step 5: Set Equal to Right-Hand Side
Now, we set this equal to the right-hand side:

\[ \tan\left(\frac{\theta}{2^1}\right) - \tan\left(\frac{\theta}{2^{2014}}\right) = \tan\left(\frac{\theta}{2^a}\right) - \tan\left(\frac{\theta}{2^b}\right). \]

### Step 6: Identify \(a\) and \(b\)
From the left-hand side, we can see that:

- The first term corresponds to \(a = 1\),
- The last term corresponds to \(b = 2014\).

### Step 7: Calculate \(a + b\)
Now, we find \(a + b\):

\[ a + b = 1 + 2014 = 2015. \]

### Final Answer
Thus, the value of \(b + a\) is:

\[ \boxed{2015}. \]