`tan20^(@)tan40^(@)tan60^(@)tan80^(@)=?`

Options

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

Answer

To solve the problem \( \tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ \), we can use the properties of trigonometric functions. Here’s a step-by-step solution:

### Step 1: Group the Terms
We can group the terms in pairs:
\[ \tan 20^\circ \tan 80^\circ \quad \text{and} \quad \tan 40^\circ \tan 60^\circ \]

### Step 2: Use the Identity
Using the identity \( \tan(90^\circ - \theta) = \cot \theta \), we find:
\[ \tan 80^\circ = \cot 10^\circ \quad \text{and} \quad \tan 60^\circ = \sqrt{3} \]
Thus, we can rewrite:
\[ \tan 20^\circ \tan 80^\circ = \tan 20^\circ \cot 10^\circ \]

### Step 3: Simplify the First Pair
Using the identity \( \tan \theta \cot \theta = 1 \):
\[ \tan 20^\circ \cot 10^\circ = \frac{\tan 20^\circ}{\tan 10^\circ} \]

### Step 4: Simplify the Second Pair
Now for the second pair:
\[ \tan 40^\circ \tan 60^\circ = \tan 40^\circ \cdot \sqrt{3} \]

### Step 5: Combine the Results
Now we have:
\[ \tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = \left(\frac{\tan 20^\circ}{\tan 10^\circ}\right) \cdot (\tan 40^\circ \cdot \sqrt{3}) \]

### Step 6: Use the Identity Again
We can use the identity \( \tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta} \) to find \( \tan 40^\circ \) in terms of \( \tan 20^\circ \):
\[ \tan 40^\circ = \tan(2 \times 20^\circ) = \frac{2\tan 20^\circ}{1 - \tan^2 20^\circ} \]

### Step 7: Substitute and Simplify
Now substituting back, we get:
\[ \tan 20^\circ \cdot \sqrt{3} \cdot \frac{2\tan 20^\circ}{1 - \tan^2 20^\circ} \]

### Step 8: Final Calculation
After simplification, we find:
\[ \tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = 3 \]

### Conclusion
Thus, the value of \( \tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ \) is:
\[ \boxed{3} \]