If `sin theta and -cos theta` are the roots of the equation `ax^(2) - bx - c = 0`, where a, b, and c are the sides of a triangle ABC, then `cos B` is equal to

Options

A. `1 - (c)/(2a)`
B. `1 - (c)/(a)`
C. `1 + (c)/(2a)`
D. `1 + (c)/(3a)`

Explanation

To solve the problem, we need to find the value of \( \cos B \) given that \( \sin \theta \) and \( -\cos \theta \) are the roots of the quadratic equation \( ax^2 - bx - c = 0 \), where \( a, b, c \) are the sides of triangle \( ABC \).

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the equation are given as \( r_1 = \sin \theta \) and \( r_2 = -\cos \theta \).

2. **Sum of the Roots**:
From Vieta's formulas, the sum of the roots \( r_1 + r_2 = \frac{b}{a} \).
\[ \sin \theta - \cos \theta = \frac{b}{a} \]

3. **Product of the Roots**:
The product of the roots \( r_1 \cdot r_2 = \frac{-c}{a} \).
\[ \sin \theta \cdot (-\cos \theta) = \frac{-c}{a} \]
This simplifies to:
\[ -\sin \theta \cos \theta = \frac{-c}{a} \implies \sin \theta \cos \theta = \frac{c}{a} \]

4. **Square the Sum of the Roots**:
We square the equation from step 2:
\[ (\sin \theta - \cos \theta)^2 = \left(\frac{b}{a}\right)^2 \]
Expanding the left-hand side:
\[ \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = \frac{b^2}{a^2} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 - 2\sin \theta \cos \theta = \frac{b^2}{a^2} \]

5. **Substituting for \( \sin \theta \cos \theta \)**:
From step 3, we know \( \sin \theta \cos \theta = \frac{c}{a} \):
\[ 1 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} \]
Rearranging gives:
\[ 1 - \frac{2c}{a} = \frac{b^2}{a^2} \]
Multiplying through by \( a^2 \):
\[ a^2 - 2ac = b^2 \]

6. **Finding \( \cos B \)**:
We use the cosine rule:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting \( b^2 = a^2 - 2ac \):
\[ \cos B = \frac{a^2 + c^2 - (a^2 - 2ac)}{2ac} \]
Simplifying:
\[ \cos B = \frac{a^2 + c^2 - a^2 + 2ac}{2ac} = \frac{c^2 + 2ac}{2ac} \]
This can be further simplified to:
\[ \cos B = \frac{c^2}{2ac} + 1 = 1 + \frac{c^2}{2ac} \]

Thus, the final answer is:
\[ \cos B = 1 + \frac{c}{2a} \]