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 \((1 + \tan \alpha)(1 + \tan 4\alpha) = 2\) given that \(0 < \alpha < \frac{\pi}{16}\), we can follow these steps:

### Step 1: Expand the equation
Start by expanding the left-hand side of the equation:
\[ (1 + \tan \alpha)(1 + \tan 4\alpha) = 1 + \tan \alpha + \tan 4\alpha + \tan \alpha \tan 4\alpha \]
Thus, we can rewrite the equation as:
\[ 1 + \tan \alpha + \tan 4\alpha + \tan \alpha \tan 4\alpha = 2 \]

### Step 2: Rearrange the equation
Now, rearranging gives us:
\[ \tan \alpha + \tan 4\alpha + \tan \alpha \tan 4\alpha = 1 \]

### Step 3: Use the tangent addition formula
Using the tangent addition formula, we know that:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Let \(A = \alpha\) and \(B = 4\alpha\). Then we have:
\[ \tan(5\alpha) = \frac{\tan \alpha + \tan 4\alpha}{1 - \tan \alpha \tan 4\alpha} \]
From our rearranged equation, we can substitute:
\[ \tan(5\alpha) = \frac{1}{1} = 1 \]

### Step 4: Solve for \(5\alpha\)
The equation \(\tan(5\alpha) = 1\) implies:
\[ 5\alpha = \frac{\pi}{4} + n\pi \quad \text{for integers } n \]
Since \(0 < \alpha < \frac{\pi}{16}\), we consider the case where \(n = 0\):
\[ 5\alpha = \frac{\pi}{4} \]

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

### Step 6: Verify the range of \(\alpha\)
Now we need to check if \(\alpha = \frac{\pi}{20}\) falls within the specified range:
\[ 0 < \frac{\pi}{20} < \frac{\pi}{16} \]
Calculating:
\[ \frac{\pi}{20} = 0.157 \quad \text{and} \quad \frac{\pi}{16} = 0.196 \]
Since \(0.157 < 0.196\), \(\alpha = \frac{\pi}{20}\) is indeed within the range.

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