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, we need to find the relationship between \( \alpha \) and \( \beta \) given the definitions of \( \alpha \) and \( \beta \).

### Step 1: Calculate \( \alpha \)

We know that:
\[ \alpha = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) + \sin^{-1}\left(\frac{1}{3}\right) \]

Using the known values:
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \]

Thus, we can rewrite \( \alpha \) as:
\[ \alpha = \frac{\pi}{3} + \sin^{-1}\left(\frac{1}{3}\right) \]

### Step 2: Calculate \( \beta \)

Next, we calculate \( \beta \):
\[ \beta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \cos^{-1}\left(\frac{1}{3}\right) \]

Using the known values:
\[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \]

Thus, we can rewrite \( \beta \) as:
\[ \beta = \frac{\pi}{6} + \cos^{-1}\left(\frac{1}{3}\right) \]

### Step 3: Compare \( \alpha \) and \( \beta \)

Now we need to compare \( \alpha \) and \( \beta \):
\[ \alpha = \frac{\pi}{3} + \sin^{-1}\left(\frac{1}{3}\right) \]
\[ \beta = \frac{\pi}{6} + \cos^{-1}\left(\frac{1}{3}\right) \]

### Step 4: Use the identity for \( \sin^{-1} \) and \( \cos^{-1} \)

We know that:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]

Thus, we can express \( \cos^{-1}\left(\frac{1}{3}\right) \) in terms of \( \sin^{-1}\left(\frac{1}{3}\right) \):
\[ \cos^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{3}\right) \]

### Step 5: Substitute back into \( \beta \)

Substituting this into \( \beta \):
\[ \beta = \frac{\pi}{6} + \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1}{3}\right)\right) \]
\[ \beta = \frac{\pi}{6} + \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{3}\right) \]
\[ \beta = \frac{\pi}{6} + \frac{3\pi}{6} - \sin^{-1}\left(\frac{1}{3}\right) \]
\[ \beta = \frac{4\pi}{6} - \sin^{-1}\left(\frac{1}{3}\right) \]
\[ \beta = \frac{2\pi}{3} - \sin^{-1}\left(\frac{1}{3}\right) \]

### Step 6: Compare \( \alpha \) and \( \beta \)

Now we have:
\[ \alpha = \frac{\pi}{3} + \sin^{-1}\left(\frac{1}{3}\right) \]
\[ \beta = \frac{2\pi}{3} - \sin^{-1}\left(\frac{1}{3}\right) \]

To compare \( \alpha \) and \( \beta \):
\[ \alpha - \beta = \left(\frac{\pi}{3} + \sin^{-1}\left(\frac{1}{3}\right)\right) - \left(\frac{2\pi}{3} - \sin^{-1}\left(\frac{1}{3}\right)\right) \]
\[ = \frac{\pi}{3} + \sin^{-1}\left(\frac{1}{3}\right) - \frac{2\pi}{3} + \sin^{-1}\left(\frac{1}{3}\right) \]
\[ = -\frac{\pi}{3} + 2\sin^{-1}\left(\frac{1}{3}\right) \]

Since \( \sin^{-1}\left(\frac{1}{3}\right) \) is positive, we can conclude that:
\[ \alpha > \beta \]

### Conclusion

Thus, we have established the relationship:
\[ \alpha > \beta \]