Consider the following statement :
1. If ABC is an equilateral triangle, then 3 tan(A + B) tan C = 1.
2. If ABC is a triangle in which `A = 78^(@), B = 66^(@)`, then `tan((A)/(2)+C) lt tan A`
3. If ABC is any triangle, then `tan((A+B)/(2))sin((C)/(2)) lt cos((C)/(2))`
Which of the above statements is/are correct?

Options

A. 1 only
B. 2 only
C. 1 and 2
D. 2 and 3

Explanation

To determine the correctness of the given statements about triangles, we will analyze each statement step by step.

### Statement 1: If ABC is an equilateral triangle, then \(3 \tan(A + B) \tan C = 1\).

1. **Understanding Equilateral Triangle**: In an equilateral triangle, all angles are equal. Thus, \(A = B = C = 60^\circ\).

2. **Calculate \(A + B\)**:
\[ A + B = 60^\circ + 60^\circ = 120^\circ \]

3. **Calculate \(\tan(A + B)\)**:
\[ \tan(120^\circ) = \tan(180^\circ - 60^\circ) = -\tan(60^\circ) = -\sqrt{3} \]

4. **Calculate \(\tan C\)**:
\[ \tan C = \tan(60^\circ) = \sqrt{3} \]

5. **Substituting into the equation**:
\[ 3 \tan(A + B) \tan C = 3 \cdot (-\sqrt{3}) \cdot (\sqrt{3}) = 3 \cdot (-3) = -9 \]

6. **Conclusion**: Since \(-9 \neq 1\), the first statement is incorrect.

### Statement 2: If \(A = 78^\circ\) and \(B = 66^\circ\), then \(\tan\left(\frac{A}{2} + C\right) < \tan A\).

1. **Calculate angle \(C\)**:
\[ C = 180^\circ - A - B = 180^\circ - 78^\circ - 66^\circ = 36^\circ \]

2. **Calculate \(\frac{A}{2}\)**:
\[ \frac{A}{2} = \frac{78^\circ}{2} = 39^\circ \]

3. **Calculate \(\tan\left(\frac{A}{2} + C\right)\)**:
\[ \tan\left(39^\circ + 36^\circ\right) = \tan(75^\circ) \]

4. **Calculate \(\tan A\)**:
\[ \tan A = \tan(78^\circ) \]

5. **Comparison**: Since \(75^\circ < 78^\circ\), it follows that \(\tan(75^\circ) < \tan(78^\circ)\).

6. **Conclusion**: The second statement is correct.

### Statement 3: If ABC is any triangle, then \(\tan\left(\frac{A + B}{2}\right) \sin\left(\frac{C}{2}\right) < \cos\left(\frac{C}{2}\right)\).

1. **Using the property of triangle**: Since \(A + B + C = 180^\circ\), we have \(A + B = 180^\circ - C\).

2. **Calculate \(\frac{A + B}{2}\)**:
\[ \frac{A + B}{2} = \frac{180^\circ - C}{2} = 90^\circ - \frac{C}{2} \]

3. **Calculate \(\tan\left(\frac{A + B}{2}\right)\)**:
\[ \tan\left(90^\circ - \frac{C}{2}\right) = \cot\left(\frac{C}{2}\right) \]

4. **Substituting into the inequality**:
\[ \cot\left(\frac{C}{2}\right) \sin\left(\frac{C}{2}\right) < \cos\left(\frac{C}{2}\right) \]

5. **Using the identity**: \(\cot(x) = \frac{\cos(x)}{\sin(x)}\), we rewrite the left side:
\[ \frac{\cos\left(\frac{C}{2}\right)}{\sin\left(\frac{C}{2}\right)} \sin\left(\frac{C}{2}\right) = \cos\left(\frac{C}{2}\right) \]

6. **Conclusion**: The inequality holds true, hence the third statement is correct.

### Final Conclusion:
- Statement 1: Incorrect
- Statement 2: Correct
- Statement 3: Correct

Thus, the correct statements are 2 and 3.