If `sin theta and cos theta` are the roots of the equation `lx^(2)+mx+n=0`, then

Options

A. `l^(2)-m^(2)+2ln=0`
B. `l^(2)+m^(2)+2ln=0`
C. `l^(2)-m^(2)-2ln=0`
D. `l^(2)+m^(2)-2ln=0`

Answer

To solve the problem, we need to find a relationship involving \( L \), \( M \), and \( N \) given that \( \sin \theta \) and \( \cos \theta \) are the roots of the quadratic equation \( Lx^2 + Mx + N = 0 \).

### Step-by-Step Solution

1. **Identify the Roots and Their Properties**:
The roots of the quadratic equation \( Lx^2 + Mx + N = 0 \) are given as \( \sin \theta \) and \( \cos \theta \).

2. **Sum of the Roots**:
According to Vieta's formulas, the sum of the roots \( \alpha + \beta \) is given by:
\[ \sin \theta + \cos \theta = -\frac{M}{L} \]

3. **Product of the Roots**:
The product of the roots \( \alpha \beta \) is given by:
\[ \sin \theta \cos \theta = \frac{N}{L} \]

4. **Use the Identity for Sum of Sines and Cosines**:
We know from trigonometric identities that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
We can also express \( \sin \theta + \cos \theta \) in terms of \( \sin^2 \theta + \cos^2 \theta \):
\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = 1 + 2\sin \theta \cos \theta \]

5. **Substituting the Values**:
Now substituting the expressions for the sum and product of the roots:
\[ \left(-\frac{M}{L}\right)^2 = 1 + 2\left(\frac{N}{L}\right) \]

6. **Simplifying the Equation**:
Expanding and rearranging gives:
\[ \frac{M^2}{L^2} = 1 + \frac{2N}{L} \]
Multiplying through by \( L^2 \):
\[ M^2 = L^2 + 2NL \]

7. **Final Rearrangement**:
Rearranging the equation gives:
\[ M^2 - 2NL - L^2 = 0 \]

### Conclusion
The relationship we derived is:
\[ M^2 - 2NL - L^2 = 0 \]