If `alpha=sin^(-1)"" 4/5+sin^(-1) ""1/3 and beta=cos^(-1) "" 4/5+cos^(-1)"" 1/3`, then

Options

A. `alpha lt beta`
B. `alpha =beta`
C. `alpha gt beta`
D. None of these

Answer

To solve the problem, we need to evaluate the expressions for \( \alpha \) and \( \beta \) and then compare them.

### Step 1: Define \( \alpha \) and \( \beta \)
We start with the definitions given in the problem:
\[ \alpha = \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{1}{3}\right) \]
\[ \beta = \cos^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{1}{3}\right) \]

### Step 2: Use the identity for \( \sin^{-1} \) and \( \cos^{-1} \)
We can use the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) to rewrite \( \alpha \):
\[ \sin^{-1}\left(\frac{4}{5}\right) = \frac{\pi}{2} - \cos^{-1}\left(\frac{4}{5}\right) \]
\[ \sin^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{3}\right) \]
Substituting these into \( \alpha \):
\[ \alpha = \left(\frac{\pi}{2} - \cos^{-1}\left(\frac{4}{5}\right)\right) + \left(\frac{\pi}{2} - \cos^{-1}\left(\frac{1}{3}\right)\right) \]
\[ \alpha = \pi - \left(\cos^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{1}{3}\right)\right) \]
Thus, we can express \( \alpha \) in terms of \( \beta \):
\[ \alpha = \pi - \beta \]

### Step 3: Compare \( \alpha \) and \( \beta \)
From the equation \( \alpha + \beta = \pi \), we can deduce:
- If \( \beta > \frac{\pi}{2} \), then \( \alpha < \frac{\pi}{2} \).
- If \( \beta < \frac{\pi}{2} \), then \( \alpha > \frac{\pi}{2} \).

### Step 4: Determine the values of \( \beta \)
Now we need to evaluate \( \beta \):
\[ \beta = \cos^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{1}{3}\right) \]

Using the formula for \( \cos^{-1}(x) + \cos^{-1}(y) \):
\[ \cos^{-1}(x) + \cos^{-1}(y) = \cos^{-1\left(xy - \sqrt{(1-x^2)(1-y^2)}\right)} \]
Let \( x = \frac{4}{5} \) and \( y = \frac{1}{3} \):
\[ \beta = \cos^{-1}\left(\frac{4}{5} \cdot \frac{1}{3} - \sqrt{\left(1 - \left(\frac{4}{5}\right)^2\right)\left(1 - \left(\frac{1}{3}\right)^2\right)}\right) \]
Calculating \( xy \):
\[ xy = \frac{4}{15} \]
Calculating \( \sqrt{(1-x^2)(1-y^2)} \):
\[ 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \quad \text{and} \quad 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]
Thus,
\[ \sqrt{\left(1 - \left(\frac{4}{5}\right)^2\right)\left(1 - \left(\frac{1}{3}\right)^2\right)} = \sqrt{\frac{9}{25} \cdot \frac{8}{9}} = \sqrt{\frac{8}{25}} = \frac{2\sqrt{2}}{5} \]
Now substituting back into the expression for \( \beta \):
\[ \beta = \cos^{-1}\left(\frac{4}{15} - \frac{2\sqrt{2}}{5}\right) \]

### Step 5: Determine the sign of \( \cos(\beta) \)
Since \( \beta \) is a sum of two \( \cos^{-1} \) values, we can analyze the behavior:
- If \( \beta > \frac{\pi}{2} \), then \( \alpha < \frac{\pi}{2} \).

### Conclusion
Since \( \alpha + \beta = \pi \), we conclude:
\[ \alpha < \beta \]

### Final Result
Thus, the final answer is:
\[ \alpha < \beta \]