In `DeltaABC,cosecA(sinBcosC+cosBsinC)=?`

Options

A. `(c )/(a)`
B. `(a)/(c )`
C. 1
D. `-1`

Answer

To solve the problem \( \csc A (\sin B \cos C + \cos B \sin C) \) in triangle \( ABC \), we can follow these steps:

### Step 1: Use the sine rule
We know from the sine rule that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
From this, we can express \( \sin B \) and \( \sin C \) in terms of \( \sin A \):
\[ \sin B = \frac{b \sin A}{a} \quad \text{and} \quad \sin C = \frac{c \sin A}{a} \]

### Step 2: Substitute \( \sin B \) and \( \sin C \)
Now, we substitute \( \sin B \) and \( \sin C \) into the expression:
\[ \sin B \cos C + \cos B \sin C = \frac{b \sin A}{a} \cos C + \cos B \frac{c \sin A}{a} \]
Factoring out \( \sin A \):
\[ = \frac{\sin A}{a} (b \cos C + c \cos B) \]

### Step 3: Substitute into the original expression
Now substitute this back into the original expression:
\[ \csc A (\sin B \cos C + \cos B \sin C) = \csc A \cdot \frac{\sin A}{a} (b \cos C + c \cos B) \]
Since \( \csc A = \frac{1}{\sin A} \), we have:
\[ = \frac{1}{\sin A} \cdot \frac{\sin A}{a} (b \cos C + c \cos B) = \frac{b \cos C + c \cos B}{a} \]

### Step 4: Use the cosine rule
Using the cosine rule, we know:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \quad \text{and} \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting these into our expression gives:
\[ = \frac{b \left( \frac{a^2 + b^2 - c^2}{2ab} \right) + c \left( \frac{a^2 + c^2 - b^2}{2ac} \right)}{a} \]

### Step 5: Simplify the expression
Now simplify the expression:
\[ = \frac{1}{2a} \left( \frac{b(a^2 + b^2 - c^2)}{ab} + \frac{c(a^2 + c^2 - b^2)}{ac} \right) \]
This simplifies to:
\[ = \frac{1}{2a} \left( \frac{a^2 + b^2 - c^2}{2} + \frac{a^2 + c^2 - b^2}{2} \right) \]

### Step 6: Final simplification
Combining the terms gives:
\[ = \frac{1}{2a} \cdot \frac{(2a^2)}{2} = 1 \]

Thus, the final answer is:
\[ \csc A (\sin B \cos C + \cos B \sin C) = 1 \]

### Conclusion
The value of \( \csc A (\sin B \cos C + \cos B \sin C) \) is \( 1 \).