if (1+tan`alpha`)(1+tan4`alpha`) =2 where `alpha` `in` (0 , `pi/16`) then `alpha ` equal to

Options

A. `(pi)/(20)`
B. `(pi)/(30)`
C. `(pi)/(40)`
D. `(pi)/(60)`

Explanation

To solve the equation \((1 + \tan \alpha)(1 + \tan 4\alpha) = 2\) where \(\alpha \in (0, \frac{\pi}{16})\), we can follow these steps:

### Step 1: Set Variables
Let \(a = \alpha\) and \(b = 4\alpha\). Then, we can rewrite the equation as:
\[ (1 + \tan a)(1 + \tan b) = 2 \]

### Step 2: Expand the Left Side
Expanding the left side gives us:
\[ 1 + \tan a + \tan b + \tan a \tan b = 2 \]
This simplifies to:
\[ \tan a + \tan b + \tan a \tan b = 1 \]

### Step 3: Rearrange the Equation
Rearranging the equation, we can write:
\[ \tan a + \tan b = 1 - \tan a \tan b \]

### Step 4: Use the Tangent Addition Formula
We know from the tangent addition formula that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
Substituting our rearranged equation into this formula gives:
\[ \tan(a + b) = 1 \]

### Step 5: Solve for \(a + b\)
Since \(\tan(a + b) = 1\), we have:
\[ a + b = \frac{\pi}{4} \]
Substituting back \(a = \alpha\) and \(b = 4\alpha\):
\[ \alpha + 4\alpha = \frac{\pi}{4} \]
This simplifies to:
\[ 5\alpha = \frac{\pi}{4} \]

### Step 6: Solve for \(\alpha\)
Dividing both sides by 5 gives:
\[ \alpha = \frac{\pi}{20} \]

### Final Result
Thus, the value of \(\alpha\) is:
\[ \alpha = \frac{\pi}{20} \]