If `tan20^0+tan40^0+tan80^0-tan60^0`= `lambdasin40^0`, find lambda.

Answer

To solve the equation \( \tan 20^\circ + \tan 40^\circ + \tan 80^\circ - \tan 60^\circ = \lambda \sin 40^\circ \), we will use a known trigonometric identity and properties of tangent.

### Step-by-Step Solution:

1. **Identify the known identity**: We know the identity for tangent:
\[ \tan A + \tan B + \tan C = \tan 3A \]
when \( A + B + C = 60^\circ \).

2. **Apply the identity**: In our case, let \( A = 20^\circ \), \( B = 40^\circ \), and \( C = 60^\circ - (20^\circ + 40^\circ) = 0^\circ \). Therefore, we can rewrite:
\[ \tan 20^\circ + \tan 40^\circ + \tan 60^\circ - \tan 60^\circ = \tan 60^\circ \]

3. **Substituting values**: We know that \( \tan 60^\circ = \sqrt{3} \). Thus, we can express the equation as:
\[ \tan 20^\circ + \tan 40^\circ + \tan 80^\circ - \sqrt{3} = 0 \]

4. **Rearranging the equation**: Rearranging gives us:
\[ \tan 20^\circ + \tan 40^\circ + \tan 80^\circ = \sqrt{3} \]

5. **Using the identity again**: We can also use the identity:
\[ \tan 20^\circ + \tan 40^\circ + \tan 80^\circ = \tan 60^\circ \]
which confirms that:
\[ \tan 20^\circ + \tan 40^\circ + \tan 80^\circ = \sqrt{3} \]

6. **Setting the equation**: Now, we have:
\[ \sqrt{3} - \sqrt{3} = \lambda \sin 40^\circ \]

7. **Simplifying the equation**: This simplifies to:
\[ 0 = \lambda \sin 40^\circ \]

8. **Finding lambda**: Since \( \sin 40^\circ \) is not zero, we can conclude:
\[ \lambda = 0 \]

### Final Answer:
\[ \lambda = 0 \]