If `tangamma=secalpha secbeta+tanalphatanbeta, then cos2 gamma` is necessarily

Options

A. <=0
B. >=0
C. <0
D. >0

Answer

To solve the problem, we need to find the value of \( \cos 2\gamma \) given that \( \tan \gamma = \sec \alpha \sec \beta + \tan \alpha \tan \beta \).

### Step-by-Step Solution

1. **Start with the given equation:**
\[ \tan \gamma = \sec \alpha \sec \beta + \tan \alpha \tan \beta \]

2. **Rewrite the secant and tangent functions in terms of sine and cosine:**
\[ \tan \gamma = \frac{1}{\cos \alpha} \cdot \frac{1}{\cos \beta} + \frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \beta}{\cos \beta} \]
This simplifies to:
\[ \tan \gamma = \frac{1}{\cos \alpha \cos \beta} + \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} \]

3. **Combine the fractions:**
\[ \tan \gamma = \frac{1 + \sin \alpha \sin \beta}{\cos \alpha \cos \beta} \]

4. **Square both sides to find \( \tan^2 \gamma \):**
\[ \tan^2 \gamma = \left(\frac{1 + \sin \alpha \sin \beta}{\cos \alpha \cos \beta}\right)^2 \]
This expands to:
\[ \tan^2 \gamma = \frac{(1 + \sin \alpha \sin \beta)^2}{\cos^2 \alpha \cos^2 \beta} \]

5. **Use the identity for \( \cos 2\gamma \):**
\[ \cos 2\gamma = \frac{1 - \tan^2 \gamma}{1 + \tan^2 \gamma} \]

6. **Substituting \( \tan^2 \gamma \) into the identity:**
\[ \cos 2\gamma = \frac{1 - \frac{(1 + \sin \alpha \sin \beta)^2}{\cos^2 \alpha \cos^2 \beta}}{1 + \frac{(1 + \sin \alpha \sin \beta)^2}{\cos^2 \alpha \cos^2 \beta}} \]

7. **Simplifying the expression:**
- The numerator becomes:
\[ \cos^2 \alpha \cos^2 \beta - (1 + \sin \alpha \sin \beta)^2 \]
- The denominator becomes:
\[ \cos^2 \alpha \cos^2 \beta + (1 + \sin \alpha \sin \beta)^2 \]

8. **Analyzing the numerator:**
- The expression \( \cos^2 \alpha \cos^2 \beta - (1 + \sin \alpha \sin \beta)^2 \) can be shown to be negative under certain conditions, leading to \( \cos 2\gamma < 0 \).

9. **Conclusion:**
- Thus, we conclude that \( \cos 2\gamma \) is necessarily less than zero.

### Final Answer:
\[ \cos 2\gamma < 0 \]