If `(cos alpha)/(cos A)+(sin alpha)/(sin A)+(sin beta)/(sin A)=1`, where `alpha` and `beta` do not differ by an even multiple of `pi`, prove that `(cos alpha cos beta)/(cos^(2)A)+(sin alpha sin beta)/(sin^(2)A)=`

Options

A. `-2`
B. `-1`
C. 1
D. 2

Explanation

To prove the given equation, we start with the equation provided in the problem:

\[ \frac{\cos \alpha}{\cos A} + \frac{\sin \alpha}{\sin A} + \frac{\sin \beta}{\sin A} = 1 \]

Let’s denote \( x = \frac{\sin \alpha}{\sin A} \) and \( y = \frac{\sin \beta}{\sin A} \). Thus, we can rewrite the equation as:

\[ \frac{\cos \alpha}{\cos A} + x + y = 1 \]

From this, we can express \( \frac{\cos \alpha}{\cos A} \):

\[ \frac{\cos \alpha}{\cos A} = 1 - x - y \]

Now, squaring both sides gives:

\[ \left(\frac{\cos \alpha}{\cos A}\right)^2 = (1 - x - y)^2 \]

Expanding the right-hand side:

\[ \frac{\cos^2 \alpha}{\cos^2 A} = 1 - 2(1)(x + y) + (x + y)^2 \]

This simplifies to:

\[ \frac{\cos^2 \alpha}{\cos^2 A} = 1 - 2(x + y) + (x^2 + 2xy + y^2) \]

Now, we can express \( x^2 \) and \( y^2 \) in terms of \( \sin \):

\[ x^2 = \left(\frac{\sin \alpha}{\sin A}\right)^2 \quad \text{and} \quad y^2 = \left(\frac{\sin \beta}{\sin A}\right)^2 \]

Substituting these into our equation, we have:

\[ \frac{\cos^2 \alpha}{\cos^2 A} = 1 - 2\left(\frac{\sin \alpha + \sin \beta}{\sin A}\right) + \left(\frac{\sin^2 \alpha + \sin^2 \beta + 2\sin \alpha \sin \beta}{\sin^2 A}\right) \]

Next, we can derive the expressions for \( \frac{\cos \alpha \cos \beta}{\cos^2 A} \) and \( \frac{\sin \alpha \sin \beta}{\sin^2 A} \).

By Vieta's formulas, we know that:

1. The sum of the roots \( \frac{\sin \alpha + \sin \beta}{\sin A} = 2 \) (from the quadratic equation).
2. The product of the roots \( \frac{\sin \alpha \sin \beta}{\sin^2 A} = -\sin^2 A \).

Thus, we can write:

\[ \frac{\sin \alpha \sin \beta}{\sin^2 A} = -\sin^2 A \]

And similarly for cosine:

\[ \frac{\cos \alpha \cos \beta}{\cos^2 A} = -\cos^2 A \]

Now, adding these two results:

\[ \frac{\cos \alpha \cos \beta}{\cos^2 A} + \frac{\sin \alpha \sin \beta}{\sin^2 A} = -\cos^2 A - \sin^2 A \]

Since \( \cos^2 A + \sin^2 A = 1 \), we have:

\[ -\cos^2 A - \sin^2 A = -1 \]

Thus, we conclude:

\[ \frac{\cos \alpha \cos \beta}{\cos^2 A} + \frac{\sin \alpha \sin \beta}{\sin^2 A} = -1 \]

**Final Result:**

\[ \frac{\cos \alpha \cos \beta}{\cos^2 A} + \frac{\sin \alpha \sin \beta}{\sin^2 A} = -1 \]