If `alpha+theta=(7pi)/(12)andtantheta=sqrt(3),` then the value of tan `alpha` is :

Options

A. `sqrt(3)`
B. 1
C. 0
D. `(1)/(sqrt(3))`

Answer

To solve the problem, we need to find the value of \( \tan \alpha \) given that \( \alpha + \theta = \frac{7\pi}{12} \) and \( \tan \theta = \sqrt{3} \).

### Step-by-step Solution:

1. **Identify \( \theta \)**:
We know that \( \tan \theta = \sqrt{3} \). The angle whose tangent is \( \sqrt{3} \) is \( \theta = \frac{\pi}{3} \) (or \( 60^\circ \)).

**Hint**: Recall the values of tangent for common angles to identify \( \theta \).

2. **Substitute \( \theta \) into the equation**:
We have \( \alpha + \theta = \frac{7\pi}{12} \). Substituting \( \theta = \frac{\pi}{3} \):
\[ \alpha + \frac{\pi}{3} = \frac{7\pi}{12} \]

**Hint**: Make sure to keep the equation balanced by substituting correctly.

3. **Solve for \( \alpha \)**:
Rearranging the equation gives:
\[ \alpha = \frac{7\pi}{12} - \frac{\pi}{3} \]
To perform the subtraction, we need a common denominator. The LCM of 12 and 3 is 12. Rewrite \( \frac{\pi}{3} \) as \( \frac{4\pi}{12} \):
\[ \alpha = \frac{7\pi}{12} - \frac{4\pi}{12} = \frac{3\pi}{12} \]
Simplifying this gives:
\[ \alpha = \frac{\pi}{4} \]

**Hint**: When subtracting fractions, always convert them to have a common denominator.

4. **Find \( \tan \alpha \)**:
Now that we have \( \alpha = \frac{\pi}{4} \), we can find \( \tan \alpha \):
\[ \tan \alpha = \tan \left( \frac{\pi}{4} \right) = 1 \]

**Hint**: Remember the special angle values for tangent to find \( \tan \alpha \).

5. **Conclusion**:
Therefore, the value of \( \tan \alpha \) is \( 1 \).

**Final Answer**: \( \tan \alpha = 1 \)