`cos(alpha+beta)+sin(alpha-beta)=0"and"tanbeta=1/2009` then value of tan3α

Options

A. 2
B. 1
C. 3
D. 4

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations:**
\[ \cos(\alpha + \beta) + \sin(\alpha - \beta) = 0 \]
\[ \tan \beta = \frac{1}{2009} \]

2. **Rewrite the first equation using trigonometric identities:**
\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]
\[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \]
Thus, we can rewrite the equation as:
\[ \cos \alpha \cos \beta - \sin \alpha \sin \beta + \sin \alpha \cos \beta - \cos \alpha \sin \beta = 0 \]

3. **Group the terms:**
\[ (\cos \alpha \cos \beta - \cos \alpha \sin \beta) + (\sin \alpha \cos \beta - \sin \alpha \sin \beta) = 0 \]
\[ \cos \alpha (\cos \beta - \sin \beta) + \sin \alpha (\cos \beta - \sin \beta) = 0 \]

4. **Factor out the common term:**
\[ (\cos \beta - \sin \beta)(\cos \alpha + \sin \alpha) = 0 \]

5. **Set each factor to zero:**
- From \(\cos \beta - \sin \beta = 0\), we get:
\[ \cos \beta = \sin \beta \implies \tan \beta = 1 \]
- From \(\cos \alpha + \sin \alpha = 0\), we get:
\[ \cos \alpha = -\sin \alpha \implies \tan \alpha = -1 \]

6. **Use the value of \(\tan \beta\):**
Since \(\tan \beta = \frac{1}{2009}\) is given, we need to use this information to find \(\tan \alpha\) correctly.

7. **Calculate \(\tan 3\alpha\):**
The formula for \(\tan 3\alpha\) is:
\[ \tan 3\alpha = \frac{3\tan \alpha - \tan^3 \alpha}{1 - 3\tan^2 \alpha} \]
Substituting \(\tan \alpha = -1\):
\[ \tan 3\alpha = \frac{3(-1) - (-1)^3}{1 - 3(-1)^2} \]
\[ = \frac{-3 + 1}{1 - 3} = \frac{-2}{-2} = 1 \]

8. **Final Answer:**
\[ \tan 3\alpha = 1 \]