If `tan20^(@)+tan40^(@)+tan80^(@)-tan60^(@)=lambda sin40^(@)`, then `lambda/4` is equal to

Answer

To solve the equation \( \tan 20^\circ + \tan 40^\circ + \tan 80^\circ - \tan 60^\circ = \lambda \sin 40^\circ \), we can follow these steps:

### Step 1: Use the formula for tangent sums
We know the identity:
\[ \tan A + \tan B + \tan C = \tan 3A \]
for \( A + B + C = 60^\circ \).

In our case, we can set:
- \( A = 20^\circ \)
- \( B = 40^\circ \)
- \( C = 60^\circ - 20^\circ - 40^\circ = 0^\circ \)

This means:
\[ \tan 20^\circ + \tan 40^\circ + \tan 0^\circ = \tan 60^\circ \]
Since \( \tan 0^\circ = 0 \), we have:
\[ \tan 20^\circ + \tan 40^\circ = \tan 60^\circ \]

### Step 2: Substitute \( \tan 60^\circ \)
We know that \( \tan 60^\circ = \sqrt{3} \). Thus:
\[ \tan 20^\circ + \tan 40^\circ = \sqrt{3} \]

### Step 3: Add \( \tan 80^\circ \)
Now, we need to add \( \tan 80^\circ \) to both sides:
\[ \tan 20^\circ + \tan 40^\circ + \tan 80^\circ = \sqrt{3} + \tan 80^\circ \]

### Step 4: Use the tangent identity
We can use the identity \( \tan(90^\circ - x) = \cot x \) to find \( \tan 80^\circ \):
\[ \tan 80^\circ = \cot 10^\circ \]
Thus, we can express \( \tan 80^\circ \) in terms of \( \tan 10^\circ \):
\[ \tan 80^\circ = \frac{1}{\tan 10^\circ} \]

### Step 5: Substitute back into the equation
Substituting back, we have:
\[ \tan 20^\circ + \tan 40^\circ + \frac{1}{\tan 10^\circ} - \sqrt{3} = \lambda \sin 40^\circ \]

### Step 6: Evaluate \( \tan 60^\circ \)
Since \( \tan 60^\circ = \sqrt{3} \), we can simplify:
\[ \tan 20^\circ + \tan 40^\circ + \tan 80^\circ - \sqrt{3} = 0 \]

### Step 7: Set the equation to zero
Thus, we have:
\[ 0 = \lambda \sin 40^\circ \]
Since \( \sin 40^\circ \neq 0 \), we conclude that \( \lambda = 0 \).

### Step 8: Find \( \lambda/4 \)
Finally, we need to find \( \frac{\lambda}{4} \):
\[ \frac{\lambda}{4} = \frac{0}{4} = 0 \]

### Final Answer:
\[ \frac{\lambda}{4} = 0 \]