If cosx.cosy + sinx.siny = -1 then cosx + cosy is

Options

A. `-2`
B. 1
C. 0
D. 2

Answer

To solve the equation \( \cos x \cos y + \sin x \sin y = -1 \), we can use the cosine addition formula. The cosine addition formula states that:

\[ \cos(x + y) = \cos x \cos y - \sin x \sin y \]

However, we can rearrange the terms in our equation to utilize the cosine addition formula. The given equation can be rewritten as:

\[ \cos x \cos y + \sin x \sin y = -1 \]

This can be recognized as:

\[ \cos(x - y) = -1 \]

From the properties of the cosine function, we know that \( \cos \theta = -1 \) when \( \theta = (2n + 1) \pi \) for any integer \( n \). Therefore, we have:

\[ x - y = (2n + 1) \pi \]

Now, we need to find \( \cos x + \cos y \). We can use the cosine addition formula again, but first, let's express \( \cos x + \cos y \) in terms of \( x - y \):

Using the identity for the sum of cosines:

\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]

We already know that \( x - y \) is an odd multiple of \( \pi \). Therefore, we can express \( \frac{x - y}{2} \) as:

\[ \frac{x - y}{2} = \frac{(2n + 1) \pi}{2} = (n + 0.5) \pi \]

This means that \( \cos\left(\frac{x - y}{2}\right) = 0 \) for any integer \( n \). Thus, we have:

\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cdot 0 = 0 \]

So, we conclude that:

\[ \cos x + \cos y = 0 \]

### Summary of Steps:
1. Start with the equation \( \cos x \cos y + \sin x \sin y = -1 \).
2. Recognize that this can be expressed as \( \cos(x - y) = -1 \).
3. Determine that \( x - y = (2n + 1) \pi \) for some integer \( n \).
4. Use the sum of cosines identity to express \( \cos x + \cos y \).
5. Conclude that \( \cos x + \cos y = 0 \).