If `0 lt alpha lt (pi)/(16) and (1+tan alpha)(1+tan4alpha)=2`, then the value of `alpha` is equal to

Options

A. `(pi)/(18)`
B. `(pi)/(20)`
C. `(pi)/(24)`
D. `(pi)/(30)`

Answer

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

### Step 1: Expand the Equation
We start by expanding the left side of the equation:
\[ (1 + \tan \alpha)(1 + \tan 4\alpha) = 1 + \tan \alpha + \tan 4\alpha + \tan \alpha \tan 4\alpha \]
Setting this equal to 2 gives us:
\[ 1 + \tan \alpha + \tan 4\alpha + \tan \alpha \tan 4\alpha = 2 \]

### Step 2: Simplify the Equation
Subtract 1 from both sides:
\[ \tan \alpha + \tan 4\alpha + \tan \alpha \tan 4\alpha = 1 \]

### Step 3: Use the Tangent Addition Formula
Recall the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
In our case, let \( A = \alpha \) and \( B = 4\alpha \). Then we can express the left side:
\[ \tan(\alpha + 4\alpha) = \tan(5\alpha) = \frac{\tan \alpha + \tan 4\alpha}{1 - \tan \alpha \tan 4\alpha} \]
From our earlier simplification, we know:
\[ \tan \alpha + \tan 4\alpha = 1 - \tan \alpha \tan 4\alpha \]
Thus, we can rewrite:
\[ \tan(5\alpha) = 1 \]

### Step 4: Solve for \( 5\alpha \)
The equation \( \tan(5\alpha) = 1 \) implies:
\[ 5\alpha = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Since \( 0 < \alpha < \frac{\pi}{16} \), we take \( n = 0 \):
\[ 5\alpha = \frac{\pi}{4} \]
Dividing both sides by 5 gives:
\[ \alpha = \frac{\pi}{20} \]

### Step 5: Verify the Condition
Now we check if \( \frac{\pi}{20} \) satisfies the condition \( 0 < \alpha < \frac{\pi}{16} \):
\[ \frac{\pi}{20} < \frac{\pi}{16} \quad \text{(since } 20 > 16\text{)} \]
Thus, the condition is satisfied.

### Final Answer
The value of \( \alpha \) is:
\[ \alpha = \frac{\pi}{20} \]