If `alpha=cos^(-1)((3)/(5)),beta=tan^(-1)((1)/(3))`, where `0ltalpha,betalt(pi)/(2)`, then `alpha-beta` is equa to :

Options

A. `tan^(-1)((p)/(5sqrt(10)))`
B. `cos^(-1)((9)/(5sqrt(10)))`
C. `tan^(-1)((9)/(13))`
D. `sin^(-1)((9)/(5sqrt(10)))`

Answer

To solve the problem where \( \alpha = \cos^{-1}\left(\frac{3}{5}\right) \) and \( \beta = \tan^{-1}\left(\frac{1}{3}\right) \), we need to find \( \alpha - \beta \).

### Step-by-Step Solution:

1. **Identify Values of Trigonometric Functions**:
- From \( \alpha = \cos^{-1}\left(\frac{3}{5}\right) \), we know that:
\[ \cos(\alpha) = \frac{3}{5} \]
- From \( \beta = \tan^{-1}\left(\frac{1}{3}\right) \), we know that:
\[ \tan(\beta) = \frac{1}{3} \]

2. **Find the Sine of Alpha**:
- We can use the Pythagorean identity to find \( \sin(\alpha) \):
\[ \sin^2(\alpha) + \cos^2(\alpha) = 1 \]
\[ \sin^2(\alpha) + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2(\alpha) + \frac{9}{25} = 1 \]
\[ \sin^2(\alpha) = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \sin(\alpha) = \frac{4}{5} \quad (\text{since } \alpha \text{ is in } (0, \frac{\pi}{2})) \]

3. **Find the Sine and Cosine of Beta**:
- Since \( \tan(\beta) = \frac{1}{3} \), we can find \( \sin(\beta) \) and \( \cos(\beta) \):
\[ \tan(\beta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{3} \]
- Let the opposite side be 1 and the adjacent side be 3. Then, the hypotenuse is:
\[ \text{hypotenuse} = \sqrt{1^2 + 3^2} = \sqrt{10} \]
- Thus,
\[ \sin(\beta) = \frac{1}{\sqrt{10}}, \quad \cos(\beta) = \frac{3}{\sqrt{10}} \]

4. **Use the Tangent Subtraction Formula**:
- The formula for \( \tan(\alpha - \beta) \) is given by:
\[ \tan(\alpha - \beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha) \tan(\beta)} \]
- First, find \( \tan(\alpha) \):
\[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]
- Now substitute \( \tan(\alpha) \) and \( \tan(\beta) \):
\[ \tan(\alpha - \beta) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{3} \cdot \frac{1}{3}} = \frac{\frac{3}{3}}{1 + \frac{4}{9}} = \frac{1}{\frac{13}{9}} = \frac{9}{13} \]

5. **Find \( \alpha - \beta \)**:
- Since \( \tan(\alpha - \beta) = \frac{9}{13} \), we can express \( \alpha - \beta \) as:
\[ \alpha - \beta = \tan^{-1}\left(\frac{9}{13}\right) \]

### Final Answer:
Thus, \( \alpha - \beta = \tan^{-1}\left(\frac{9}{13}\right) \).