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)/(14))`
D. `sin^(-1)((9)/(5sqrt(10)))`

Answer

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

### Step 1: Find \( \tan(\alpha) \)
Since \( \alpha = \cos^{-1}\left(\frac{3}{5}\right) \), we can find \( \tan(\alpha) \) using the relationship between sine, cosine, and tangent.

1. From \( \cos(\alpha) = \frac{3}{5} \), we can find \( \sin(\alpha) \) using the Pythagorean identity:
\[ \sin^2(\alpha) + \cos^2(\alpha) = 1 \]
\[ \sin^2(\alpha) + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2(\alpha) + \frac{9}{25} = 1 \]
\[ \sin^2(\alpha) = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \sin(\alpha) = \frac{4}{5} \]

2. Now, we can find \( \tan(\alpha) \):
\[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

### Step 2: Find \( \tan(\beta) \)
Given \( \beta = \tan^{-1}\left(\frac{1}{3}\right) \), we already know:
\[ \tan(\beta) = \frac{1}{3} \]

### Step 3: Use the tangent subtraction formula
We can now use the tangent subtraction formula to find \( \tan(\alpha - \beta) \):
\[ \tan(\alpha - \beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha) \tan(\beta)} \]
Substituting the values we found:
\[ \tan(\alpha - \beta) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + \left(\frac{4}{3}\right)\left(\frac{1}{3}\right)} \]
\[ = \frac{\frac{3}{3}}{1 + \frac{4}{9}} = \frac{1}{\frac{13}{9}} = \frac{9}{13} \]

### Step 4: Find \( \alpha - \beta \)
Now we have \( \tan(\alpha - \beta) = \frac{9}{13} \). To find \( \alpha - \beta \), we can express it as:
\[ \alpha - \beta = \tan^{-1}\left(\frac{9}{13}\right) \]

### Final Answer
Thus, the value of \( \alpha - \beta \) is:
\[ \alpha - \beta = \tan^{-1}\left(\frac{9}{13}\right) \]