If `alpha = sin^(-1)(sqrt(3)/2)+sin^(-1)(1/3) , beta =cos ^(-1)(sqrt(3)/2)+cos^(-1)(1/3)` then

Options

A. `alpha gt beta`
B. `alpha = beta`
C. `alpha lt beta`
D. `alpha + beta =2pi`

Answer

To solve the problem given, we need to evaluate the expressions for \( \alpha \) and \( \beta \) and then find their relationship.

### Step 1: Evaluate \( \alpha \)
Given:
\[ \alpha = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) + \sin^{-1}\left(\frac{1}{3}\right) \]

From trigonometric values, we know:
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \]

Let:
\[ y = \sin^{-1}\left(\frac{1}{3}\right) \]

Thus, we can rewrite \( \alpha \) as:
\[ \alpha = \frac{\pi}{3} + y \]

### Step 2: Evaluate \( \beta \)
Given:
\[ \beta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \cos^{-1}\left(\frac{1}{3}\right) \]

From trigonometric values, we know:
\[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \]

Let:
\[ z = \cos^{-1}\left(\frac{1}{3}\right) \]

Thus, we can rewrite \( \beta \) as:
\[ \beta = \frac{\pi}{6} + z \]

### Step 3: Relationship between \( \alpha \) and \( \beta \)
Now, we need to find the relationship between \( \alpha \) and \( \beta \):
\[ \alpha = \frac{\pi}{3} + y \]
\[ \beta = \frac{\pi}{6} + z \]

### Step 4: Find \( \alpha + \beta \)
Adding \( \alpha \) and \( \beta \):
\[ \alpha + \beta = \left(\frac{\pi}{3} + y\right) + \left(\frac{\pi}{6} + z\right) \]

To add \( \frac{\pi}{3} \) and \( \frac{\pi}{6} \), we find a common denominator:
\[ \frac{\pi}{3} = \frac{2\pi}{6} \]
Thus:
\[ \alpha + \beta = \frac{2\pi}{6} + \frac{\pi}{6} + y + z = \frac{3\pi}{6} + y + z = \frac{\pi}{2} + y + z \]

### Step 5: Evaluate \( y + z \)
Since \( y = \sin^{-1}\left(\frac{1}{3}\right) \) and \( z = \cos^{-1}\left(\frac{1}{3}\right) \), we know:
\[ y + z = \frac{\pi}{2} \]

### Step 6: Substitute back
Substituting back into the equation:
\[ \alpha + \beta = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

### Conclusion
Thus, we have:
\[ \alpha + \beta = \pi \]

### Final Answer
The final answer is:
\[ \alpha + \beta = \pi \]