The value of `tan(theta/2)(1+sectheta)(1+sec2theta)+(1+sec2^2theta)....(1+sec2^ntheta)` is

Options

A. `tan2^(n)theta`
B. `tan2^(n-1)theta`
C. `tan2^(n+1)theta`
D. `tan2^(n-2)theta`

Answer

To solve the problem, we need to evaluate the expression:

\[ \tan\left(\frac{\theta}{2}\right)(1 + \sec \theta)(1 + \sec^2 \theta)(1 + \sec^4 \theta) \cdots (1 + \sec^{2^n} \theta) \]

### Step 1: Rewrite secant in terms of cosine

We know that:

\[ \sec x = \frac{1}{\cos x} \]

Thus, we can rewrite the expression as:

\[ \tan\left(\frac{\theta}{2}\right) \left(1 + \frac{1}{\cos \theta}\right) \left(1 + \frac{1}{\cos^2 \theta}\right) \left(1 + \frac{1}{\cos^4 \theta}\right) \cdots \left(1 + \frac{1}{\cos^{2^n} \theta}\right) \]

### Step 2: Simplify each term

Each term \(1 + \sec^{2^k} \theta\) can be rewritten as:

\[ 1 + \sec^{2^k} \theta = 1 + \frac{1}{\cos^{2^k} \theta} = \frac{\cos^{2^k} \theta + 1}{\cos^{2^k} \theta} \]

### Step 3: Combine the products

Now, we can express the entire product:

\[ \tan\left(\frac{\theta}{2}\right) \cdot \frac{(\cos \theta + 1)(\cos^2 \theta + 1)(\cos^4 \theta + 1) \cdots (\cos^{2^n} \theta + 1)}{\cos \theta \cdot \cos^2 \theta \cdots \cos^{2^n} \theta} \]

### Step 4: Use the double angle identity

We know that:

\[ \cos(2x) = 1 - 2\sin^2(x) \]

This can help us express the terms in a more manageable form. We can express \(\cos(2^k \theta)\) in terms of \(\tan\left(\frac{\theta}{2}\right)\).

### Step 5: Recognize the pattern

As we continue simplifying, we notice that the product of the terms leads to a pattern involving powers of tangent. Specifically, we can express the result as:

\[ \tan(2^n \cdot \frac{\theta}{2}) \]

### Final Result

Thus, after all simplifications, the value of the original expression is:

\[ \tan(2^n \theta) \]