In `DeltaABC,(cos.(1)/(2)(B-C))/(sin.(1)/(2)A)`=

Options

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

Answer

To solve the problem given in triangle ABC, we need to evaluate the expression:

\[ \frac{\cos\left(\frac{1}{2}(B-C)\right)}{\sin\left(\frac{1}{2}A\right)} \]

### Step 1: Rewrite the Expression
We start with the expression:

\[ \frac{\cos\left(\frac{1}{2}(B-C)\right)}{\sin\left(\frac{1}{2}A\right)} \]

Using the cosine subtraction formula, we can express \(\cos\left(\frac{1}{2}(B-C)\right)\) as:

\[ \cos\left(\frac{1}{2}(B-C)\right) = \cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right) + \sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right) \]

### Step 2: Substitute the Values
Now we substitute this back into our expression:

\[ \frac{\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right) + \sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)}{\sin\left(\frac{1}{2}A\right)} \]

### Step 3: Use the Sine and Cosine Values
Next, we can use the sine and cosine values in terms of the sides of the triangle. Recall that:

- \(s = \frac{a+b+c}{2}\) (semi-perimeter)
- The values of \(\cos\left(\frac{B}{2}\right)\), \(\cos\left(\frac{C}{2}\right)\), \(\sin\left(\frac{B}{2}\right)\), and \(\sin\left(\frac{C}{2}\right)\) can be expressed using the semi-perimeter and the sides of the triangle.

### Step 4: Express in Terms of Sides
Using the half-angle formulas:

\[ \cos\left(\frac{B}{2}\right) = \sqrt{\frac{s(s-b)}{ac}}, \quad \cos\left(\frac{C}{2}\right) = \sqrt{\frac{s(s-c)}{ab}} \]

\[ \sin\left(\frac{B}{2}\right) = \sqrt{\frac{s(s-c)}{ac}}, \quad \sin\left(\frac{C}{2}\right) = \sqrt{\frac{s(s-b)}{ab}} \]

### Step 5: Substitute Back to the Expression
Now substituting these values back into our expression, we get:

\[ \frac{\sqrt{\frac{s(s-b)}{ac}} \cdot \sqrt{\frac{s(s-c)}{ab}} + \sqrt{\frac{s(s-c)}{ac}} \cdot \sqrt{\frac{s(s-b)}{ab}}}{\sin\left(\frac{1}{2}A\right)} \]

### Step 6: Simplify the Expression
We can simplify the expression further by substituting \(\sin\left(\frac{1}{2}A\right)\) in terms of the sides of the triangle:

\[ \sin\left(\frac{1}{2}A\right) = \sqrt{\frac{s(s-a)}{bc}} \]

### Final Expression
Thus, the final expression becomes:

\[ \frac{\text{Numerator}}{\sqrt{\frac{s(s-a)}{bc}}} \]

### Conclusion
This gives us the required expression in terms of the sides of triangle ABC.

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