Find the sum n terms of seires `tan theta+1/2tan(theta/2 )+1/(2^2)tan(theta/(2^2))+1/(2^3)tan(theta/(2^3))+....`

Options

A. `theta - cot 2theta`
B. `1/theta - 2 cot 2 theta`
C. `theta - 1/2 cot 2 theta`
D. `1/theta - 1/2 cot theta`

Answer

To find the sum of the series

\[ S_n = \tan \theta + \frac{1}{2} \tan \left( \frac{\theta}{2} \right) + \frac{1}{2^2} \tan \left( \frac{\theta}{2^2} \right) + \frac{1}{2^3} \tan \left( \frac{\theta}{2^3} \right) + \ldots \]

up to \( n \) terms, we can follow these steps:

### Step 1: Identify the general term of the series
The general term of the series can be expressed as:

\[ T_k = \frac{1}{2^{k-1}} \tan \left( \frac{\theta}{2^{k-1}} \right) \]

for \( k = 1, 2, \ldots, n \).

### Step 2: Write the sum of the series
The sum of the first \( n \) terms can be written as:

\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1}{2^{k-1}} \tan \left( \frac{\theta}{2^{k-1}} \right) \]

### Step 3: Use the identity for tangent
We know that:

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

This can help us express \( \tan \left( \frac{\theta}{2^k} \right) \) in terms of \( \tan \theta \) and \( \tan \left( \frac{\theta}{2^{k-1}} \right) \).

### Step 4: Express the series in a simplified form
By applying the tangent identity recursively, we can express each term in terms of \( \tan \theta \) and \( \tan \left( \frac{\theta}{2^{k-1}} \right) \).

### Step 5: Sum the series
After simplifying, we find that the sum can be expressed as:

\[ S_n = \tan \theta - \cos \theta + \frac{1}{2^n} \cos \left( \frac{\theta}{2^n} \right) \]

### Final Result
Thus, the sum of the first \( n \) terms of the series is:

\[ S_n = \tan \theta - \cos \theta + \frac{1}{2^n} \cos \left( \frac{\theta}{2^n} \right) \]