Consider the following for triangle ABC
I. `Sin((B+C)/(2))=Cos((A)/(2))`
II. `tan((B+C)/(2))=Cot((A)/(2))`
III. `Sin (b+c) = Cos A`
IV. `tan(B+C)=-CotA`
Which of the above ar correct?

Options

A. I and III
B. I and II
C. I and IV
D. II and III

Answer

To determine which of the statements for triangle ABC are correct, we will analyze each statement step by step.

### Step 1: Analyze Statement I
**Statement I:** \( \sin\left(\frac{B+C}{2}\right) = \cos\left(\frac{A}{2}\right) \)

Using the property of angles in a triangle, we know that \( B + C = 180^\circ - A \). Therefore, we can rewrite the left side:
\[ \frac{B+C}{2} = \frac{180^\circ - A}{2} = 90^\circ - \frac{A}{2} \]
Thus,
\[ \sin\left(\frac{B+C}{2}\right) = \sin\left(90^\circ - \frac{A}{2}\right) = \cos\left(\frac{A}{2}\right) \]
This confirms that Statement I is **true**.

### Step 2: Analyze Statement II
**Statement II:** \( \tan\left(\frac{B+C}{2}\right) = \cot\left(\frac{A}{2}\right) \)

From the previous calculation, we have:
\[ \frac{B+C}{2} = 90^\circ - \frac{A}{2} \]
Thus,
\[ \tan\left(\frac{B+C}{2}\right) = \tan\left(90^\circ - \frac{A}{2}\right) = \cot\left(\frac{A}{2}\right) \]
This confirms that Statement II is also **true**.

### Step 3: Analyze Statement III
**Statement III:** \( \sin(B+C) = \cos A \)

We know that \( B + C = 180^\circ - A \), so:
\[ \sin(B+C) = \sin(180^\circ - A) = \sin A \]
However, \( \cos A \) is not equal to \( \sin A \) in general. Therefore, Statement III is **false**.

### Step 4: Analyze Statement IV
**Statement IV:** \( \tan(B+C) = -\cot A \)

Again, using \( B + C = 180^\circ - A \):
\[ \tan(B+C) = \tan(180^\circ - A) = -\tan A \]
And since \( \cot A = \frac{1}{\tan A} \), we have:
\[ -\cot A = -\frac{1}{\tan A} \]
Thus, Statement IV is also **false**.

### Conclusion
The correct statements are:
- Statement I: True
- Statement II: True
- Statement III: False
- Statement IV: False

Thus, the answer is **Option B: 1st and 2nd**.