If sin `theta ` and - cos `theta ` are the roots of the equation `ax^2-bx-c=0` where a, b and c the side 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)`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Identify the roots of the equation
We are given that the roots of the quadratic equation \( ax^2 - bx - c = 0 \) are \( \sin \theta \) and \( -\cos \theta \).

### Step 2: Use the sum of the roots
The sum of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by the formula:
\[ \text{Sum of roots} = -\frac{b}{a} \]
For our equation, the sum of the roots \( \sin \theta + (-\cos \theta) = \sin \theta - \cos \theta \). Therefore, we have:
\[ \sin \theta - \cos \theta = -\frac{-b}{a} = \frac{b}{a} \]

### Step 3: Use the product of the roots
The product of the roots is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
So, we have:
\[ \sin \theta \cdot (-\cos \theta) = -\sin \theta \cos \theta = \frac{c}{a} \]
This implies:
\[ \sin \theta \cos \theta = -\frac{c}{a} \]

### Step 4: Square the difference of the roots
Now, we can square the difference of the roots:
\[ (\sin \theta - \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we get:
\[ \left(\frac{b}{a}\right)^2 = 1 - 2\left(-\frac{c}{a}\right) \]
This simplifies to:
\[ \frac{b^2}{a^2} = 1 + \frac{2c}{a} \]

### Step 5: Rearranging the equation
Multiplying through by \( a^2 \) gives:
\[ b^2 = a^2 + 2ac \]

### Step 6: Use the cosine rule
From the cosine rule in triangle ABC, we know:
\[ b^2 = a^2 + c^2 - 2ac \cos B \]
Setting the two expressions for \( b^2 \) equal to each other:
\[ a^2 + 2ac = a^2 + c^2 - 2ac \cos B \]
Cancelling \( a^2 \) from both sides gives:
\[ 2ac = c^2 - 2ac \cos B \]

### Step 7: Solve for \( \cos B \)
Rearranging gives:
\[ 2ac \cos B = c^2 - 2ac \]
Thus:
\[ \cos B = \frac{c^2 - 2ac}{2ac} \]

### Step 8: Simplifying the expression
Factoring out \( c \) from the numerator:
\[ \cos B = \frac{c(c - 2a)}{2ac} = \frac{c - 2a}{2a} \]

### Final Result
The value of \( \cos B \) is:
\[ \cos B = 1 + \frac{c}{2a} \]