`tan""(pi)/(8)tan""(pi)/(12)tan""(3pi)/(8)tan""(5pi)/(12)-sin^(2)""(pi)/(6)=?`

Options

A. `(1)/(2)`
B. `(2-sqrt(3))/(2)`
C. `(1)/(4)`
D. `(3)/(4)`

Answer

To solve the expression \( \tan\left(\frac{\pi}{8}\right) \tan\left(\frac{\pi}{12}\right) \tan\left(\frac{3\pi}{8}\right) \tan\left(\frac{5\pi}{12}\right) - \sin^2\left(\frac{\pi}{6}\right) \), we can follow these steps:

### Step 1: Calculate \( \sin^2\left(\frac{\pi}{6}\right) \)
We know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Thus,
\[ \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 2: Use the identity for tangent products
We can use the identity:
\[ \tan(a) \tan(b) = 1 \quad \text{if } a + b = \frac{\pi}{2} \]
First, we will pair the tangents:
- \( a = \frac{\pi}{8} \) and \( b = \frac{3\pi}{8} \)
- \( a = \frac{\pi}{12} \) and \( b = \frac{5\pi}{12} \)

### Step 3: Check the first pair
For the first pair:
\[ \frac{\pi}{8} + \frac{3\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2} \]
Thus,
\[ \tan\left(\frac{\pi}{8}\right) \tan\left(\frac{3\pi}{8}\right) = 1 \]

### Step 4: Check the second pair
For the second pair:
\[ \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2} \]
Thus,
\[ \tan\left(\frac{\pi}{12}\right) \tan\left(\frac{5\pi}{12}\right) = 1 \]

### Step 5: Combine the results
Now we can combine the results:
\[ \tan\left(\frac{\pi}{8}\right) \tan\left(\frac{3\pi}{8}\right) \tan\left(\frac{\pi}{12}\right) \tan\left(\frac{5\pi}{12}\right) = 1 \cdot 1 = 1 \]

### Step 6: Substitute back into the equation
Now substituting back into the original expression:
\[ 1 - \sin^2\left(\frac{\pi}{6}\right) = 1 - \frac{1}{4} \]

### Step 7: Calculate the final result
Calculating this gives:
\[ 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \]

### Final Answer
Thus, the final result is:
\[ \frac{3}{4} \]

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