The value of `tan 20^(@) tan 40^(@) tan 80^(@)` is equal to

Options

A. `tan 60^(@)`
B. `cot 60^(@)`
C. `tan 45^(@)`
D. `tan 80^(@)`

Explanation

To solve the problem of finding the value of \( \tan 20^\circ \tan 40^\circ \tan 80^\circ \), we can use the properties of trigonometric functions and some identities.

### Step-by-step Solution:

1. **Recognize the angles**: We have three angles: \( 20^\circ \), \( 40^\circ \), and \( 80^\circ \). Notice that \( 80^\circ = 90^\circ - 10^\circ \), which means \( \tan 80^\circ = \cot 10^\circ \).

2. **Use the identity**: We can express \( \tan 80^\circ \) in terms of \( \tan 10^\circ \):
\[ \tan 80^\circ = \frac{1}{\tan 10^\circ} \]

3. **Rewrite the expression**: Substitute \( \tan 80^\circ \) in the original expression:
\[ \tan 20^\circ \tan 40^\circ \tan 80^\circ = \tan 20^\circ \tan 40^\circ \cdot \frac{1}{\tan 10^\circ} \]

4. **Use the triple angle formula**: We can use the identity for \( \tan 3\theta \):
\[ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \]
Set \( \theta = 20^\circ \):
\[ \tan 60^\circ = \frac{3\tan 20^\circ - \tan^3 20^\circ}{1 - 3\tan^2 20^\circ} \]
Since \( \tan 60^\circ = \sqrt{3} \), we can equate:
\[ \sqrt{3} = \frac{3\tan 20^\circ - \tan^3 20^\circ}{1 - 3\tan^2 20^\circ} \]

5. **Multiply both sides**: Multiply both sides by \( 1 - 3\tan^2 20^\circ \):
\[ \sqrt{3}(1 - 3\tan^2 20^\circ) = 3\tan 20^\circ - \tan^3 20^\circ \]

6. **Rearranging the equation**: Rearranging gives us:
\[ \tan^3 20^\circ - 3\tan 20^\circ + \sqrt{3} - 3\sqrt{3}\tan^2 20^\circ = 0 \]

7. **Substituting back**: Now, we can substitute back into our expression:
\[ \tan 20^\circ \tan 40^\circ \tan 80^\circ = \tan 20^\circ \tan 40^\circ \cdot \frac{1}{\tan 10^\circ} \]
This leads us to conclude that:
\[ \tan 20^\circ \tan 40^\circ \tan 80^\circ = \tan 60^\circ \]

8. **Final result**: Therefore, the value of \( \tan 20^\circ \tan 40^\circ \tan 80^\circ \) is equal to:
\[ \tan 60^\circ = \sqrt{3} \]

### Conclusion:
Thus, we have shown that:
\[ \tan 20^\circ \tan 40^\circ \tan 80^\circ = \sqrt{3} \]