Let `(1+tan1^(ul0))(1+tan2^(ul0))......(1+tan45^(ul0))=2^(k)` then `k` equal to

Options

A. `21`
B. `22`
C. `23`
D. `24`

Answer

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

\[ (1 + \tan 1^\circ)(1 + \tan 2^\circ)(1 + \tan 3^\circ) \ldots (1 + \tan 45^\circ) \]

Let's denote this product as \( S \).

### Step 1: Rewrite the product

We can express \( S \) as:

\[ S = \prod_{k=1}^{45} (1 + \tan k^\circ) \]

### Step 2: Use the identity for tangent

We know that:

\[ \tan(45^\circ - k^\circ) = \frac{1 - \tan k^\circ}{1 + \tan k^\circ} \]

This means that:

\[ 1 + \tan(45^\circ - k^\circ) = 1 + \frac{1 - \tan k^\circ}{1 + \tan k^\circ} = \frac{(1 + \tan k^\circ) + (1 - \tan k^\circ)}{1 + \tan k^\circ} = \frac{2}{1 + \tan k^\circ} \]

### Step 3: Pair the terms

Now we can pair terms in the product \( S \):

\[ S = (1 + \tan 1^\circ)(1 + \tan 44^\circ) \cdots (1 + \tan 22^\circ)(1 + \tan 23^\circ) \]

For each pair \( (1 + \tan k^\circ)(1 + \tan(45^\circ - k^\circ)) \):

\[ (1 + \tan k^\circ)(1 + \tan(45^\circ - k^\circ) = (1 + \tan k^\circ) \cdot \frac{2}{1 + \tan k^\circ} = 2 \]

### Step 4: Count the pairs

Since we have pairs from \( k = 1 \) to \( k = 22 \) and the middle term \( (1 + \tan 45^\circ) \):

- There are \( 22 \) pairs contributing \( 2 \) each.
- The term \( (1 + \tan 45^\circ) = 1 + 1 = 2 \).

Thus, we have:

\[ S = 2^{22} \cdot 2 = 2^{23} \]

### Step 5: Find \( k \)

From the expression \( S = 2^k \), we see that:

\[ k = 23 \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{23} \]