In a tringle `ABC, sin A-cosB=cosC,` then angle B, is

Options

A. `pi//2`
B. `pi//3`
C. `pi//4`
D. `pi//6`

Answer

To solve the problem, we need to find angle B in triangle ABC given the equation:

\[ \sin A - \cos B = \cos C \]

### Step 1: Use the triangle angle sum property
In any triangle, the sum of the angles is equal to \(180^\circ\) or \(\pi\) radians. Therefore, we can write:

\[ A + B + C = \pi \]

### Step 2: Rearranging the given equation
We can rearrange the given equation to express \(\sin A\):

\[ \sin A = \cos B + \cos C \]

### Step 3: Use the sine and cosine formulas
Using the sine double angle formula, we can express \(\sin A\):

\[ \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2} \]

For \(\cos B + \cos C\), we can use the cosine addition formula:

\[ \cos B + \cos C = 2 \cos \left(\frac{B + C}{2}\right) \cos \left(\frac{B - C}{2}\right) \]

Since \(B + C = \pi - A\), we can write:

\[ \cos B + \cos C = 2 \cos \left(\frac{\pi - A}{2}\right) \cos \left(\frac{B - C}{2}\right) \]

### Step 4: Simplifying the cosine expression
Using the identity \(\cos(\frac{\pi - A}{2}) = \sin(\frac{A}{2})\), we have:

\[ \cos B + \cos C = 2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{B - C}{2}\right) \]

### Step 5: Equating the two expressions
Now we can equate the two expressions for \(\sin A\):

\[ 2 \sin \frac{A}{2} \cos \frac{A}{2} = 2 \sin \frac{A}{2} \cos \left(\frac{B - C}{2}\right) \]

### Step 6: Canceling common terms
Assuming \(\sin \frac{A}{2} \neq 0\), we can divide both sides by \(2 \sin \frac{A}{2}\):

\[ \cos \frac{A}{2} = \cos \left(\frac{B - C}{2}\right) \]

### Step 7: Using the cosine identity
From the cosine identity, we know that:

\[ \frac{A}{2} = \frac{B - C}{2} \quad \text{or} \quad \frac{A}{2} = -\frac{B - C}{2} \]

This gives us two cases to consider.

### Step 8: Solving the first case
For the first case:

\[ A = B - C \]

Substituting this into the angle sum property:

\[ (B - C) + B + C = \pi \]

This simplifies to:

\[ 2B = \pi \quad \Rightarrow \quad B = \frac{\pi}{2} \]

### Step 9: Solving the second case
For the second case:

\[ A = C - B \]

Substituting this into the angle sum property:

\[ (C - B) + B + C = \pi \]

This simplifies to:

\[ 2C = \pi \quad \Rightarrow \quad C = \frac{\pi}{2} \]

However, since we are looking for angle B, we conclude that:

\[ B = \frac{\pi}{2} \]

### Final Answer
Thus, the angle B is:

\[ \boxed{\frac{\pi}{2}} \text{ or } 90^\circ \]