If `x_k = (sectheta)^(1/2^k)+ (tan theta)^(1/(2^k) ` `and y_k = (sectheta)^(1/(2^k))-(tan theta)^(1/2^k)` , then value of `3y_n prod_(k=0)^n (x_k)` is equal to

Answer

To solve the problem, we need to evaluate the expression \( 3y_n \prod_{k=0}^{n} x_k \) where:

\[ x_k = (\sec \theta)^{1/2^k} + (\tan \theta)^{1/2^k} \]
\[ y_k = (\sec \theta)^{1/2^k} - (\tan \theta)^{1/2^k} \]

### Step 1: Express \( y_n \)

We can express \( y_n \) as:

\[ y_n = (\sec \theta)^{1/2^n} - (\tan \theta)^{1/2^n} \]

### Step 2: Express \( \prod_{k=0}^{n} x_k \)

Next, we need to evaluate the product \( \prod_{k=0}^{n} x_k \):

\[ x_k = (\sec \theta)^{1/2^k} + (\tan \theta)^{1/2^k} \]

Using the identity \( a^2 - b^2 = (a-b)(a+b) \), we can rewrite the product:

\[ \prod_{k=0}^{n} x_k = \prod_{k=0}^{n} \left( (\sec \theta)^{1/2^k} + (\tan \theta)^{1/2^k} \right) \]

### Step 3: Use the identity for \( x_k \) and \( y_k \)

Notice that:

\[ x_k y_k = \left( (\sec \theta)^{1/2^k} + (\tan \theta)^{1/2^k} \right) \left( (\sec \theta)^{1/2^k} - (\tan \theta)^{1/2^k} \right) = (\sec \theta)^{2/2^k} - (\tan \theta)^{2/2^k} = \sec^{2/2^k} \theta - \tan^{2/2^k} \theta \]

### Step 4: Simplify \( \prod_{k=0}^{n} x_k \)

The product can be simplified using the above identity:

\[ \prod_{k=0}^{n} x_k = \prod_{k=0}^{n} \left( y_k + 2(\tan \theta)^{1/2^k} \right) \]

### Step 5: Combine \( 3y_n \) and \( \prod_{k=0}^{n} x_k \)

Now, we can combine \( 3y_n \) with the product:

\[ 3y_n \prod_{k=0}^{n} x_k = 3 \left( (\sec \theta)^{1/2^n} - (\tan \theta)^{1/2^n} \right) \prod_{k=0}^{n} x_k \]

### Step 6: Evaluate the final expression

Using the identity \( \sec^2 \theta - \tan^2 \theta = 1 \):

\[ 3y_n \prod_{k=0}^{n} x_k = 3 \cdot 1 = 3 \]

### Final Answer

Thus, the value of \( 3y_n \prod_{k=0}^{n} x_k \) is:

\[ \boxed{3} \]