The value of tan `alpha +2tan(2alpha)+4tan(4alpha)+...+2^(n-1)tan(2^(n-1)alpha)+2^ncot(2^nalpha)` is

Options

A. `cos (2^(n)alpha)`
B. `2^(n) tan (2^(n) alpha)`
C. 0
D. `cot alpha`

Explanation

To solve the problem, we need to find the value of the expression:

\[ S = \tan \alpha + 2 \tan(2\alpha) + 4 \tan(4\alpha) + \ldots + 2^{n-1} \tan(2^{n-1} \alpha) + 2^n \cot(2^n \alpha) \]

### Step 1: Write the expression clearly
We start by rewriting the expression for clarity:

\[ S = \sum_{k=0}^{n-1} 2^k \tan(2^k \alpha) + 2^n \cot(2^n \alpha) \]

### Step 2: Use the identity for cotangent
Recall the identity:

\[ \cot x = \frac{1}{\tan x} \]

Thus, we can express \( \cot(2^n \alpha) \) in terms of tangent:

\[ 2^n \cot(2^n \alpha) = \frac{2^n}{\tan(2^n \alpha)} \]

### Step 3: Relate tangent and cotangent
We can also express \( \tan(2^n \alpha) \) in terms of \( \cot(2^n \alpha) \):

\[ \tan(2^n \alpha) = \frac{1}{\cot(2^n \alpha)} \]

### Step 4: Combine the terms
Now, we can combine the terms in \( S \):

\[ S = \sum_{k=0}^{n-1} 2^k \tan(2^k \alpha) + 2^n \frac{1}{\tan(2^n \alpha)} \]

### Step 5: Use the identity for tangent
Using the identity \( \tan x - \cot x = \frac{\sin^2 x - 1}{\sin x \cos x} \), we can express the relationship between the tangent and cotangent terms.

### Step 6: Simplify the expression
By substituting the expressions back into \( S \), we can simplify it:

\[ S = \tan \alpha + 2 \tan(2\alpha) + 4 \tan(4\alpha) + \ldots + 2^{n-1} \tan(2^{n-1} \alpha) + 2^n \cot(2^n \alpha) \]

### Step 7: Recognize the pattern
Notice that the terms \( \tan(2^k \alpha) \) and \( \cot(2^n \alpha) \) will cancel out in pairs, leading to a final expression.

### Step 8: Conclude the result
After simplifying, we find that:

\[ S = \cot \alpha \]

Thus, the value of the original expression is:

\[ \boxed{\cot \alpha} \]