If `(1+tan1^(@)) *(1+tan2^(@))*(1+tan3^(@)) …….(1+tan45^(@)) = 2^(n)`, then 'n' is equal to :

Options

A. 16
B. 23
C. 30
D. None of these

Answer

To solve the equation \((1 + \tan 1^\circ)(1 + \tan 2^\circ)(1 + \tan 3^\circ) \ldots (1 + \tan 45^\circ) = 2^n\), we will follow these steps:

### Step 1: Pairing the Terms
We can pair the terms in the product as follows:
\[ (1 + \tan x)(1 + \tan(45^\circ - x)) \]
for \(x = 1^\circ, 2^\circ, \ldots, 22^\circ\).

### Step 2: Simplifying Each Pair
Using the identity \(\tan(45^\circ - x) = \frac{1 - \tan x}{1 + \tan x}\), we can simplify:
\[ 1 + \tan(45^\circ - x) = 1 + \frac{1 - \tan x}{1 + \tan x} \]
Now, simplifying the expression:
\[ 1 + \tan(45^\circ - x) = \frac{(1 + \tan x) + (1 - \tan x)}{1 + \tan x} = \frac{2}{1 + \tan x} \]

### Step 3: Combining the Terms
Thus, we have:
\[ (1 + \tan x)(1 + \tan(45^\circ - x)) = (1 + \tan x) \cdot \frac{2}{1 + \tan x} = 2 \]

### Step 4: Counting the Pairs
Since we can form pairs from \(1^\circ\) to \(44^\circ\), we have:
- Pairs: \((1^\circ, 44^\circ), (2^\circ, 43^\circ), \ldots, (22^\circ, 23^\circ)\)
- This gives us \(22\) pairs, each contributing a factor of \(2\).

### Step 5: Including \(\tan 45^\circ\)
Now, we also have the term \(1 + \tan 45^\circ\):
\[ 1 + \tan 45^\circ = 1 + 1 = 2 \]

### Step 6: Total Contribution
The total contribution from all pairs and the term for \(45^\circ\) is:
\[ 2^{22} \cdot 2 = 2^{23} \]

### Step 7: Conclusion
Thus, we can equate this to the given expression:
\[ 2^{23} = 2^n \]
From this, we find:
\[ n = 23 \]

### Final Answer
Therefore, the value of \(n\) is \(23\).

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