If `alpha = cos^(-1)((3)/(5)), beta = tan ^(-1)((1)/(3))` , where `0 lt alpha, beta lt (pi)/(2)`, then `alpha - beta` is equal to

Options

A. `cos^(-1)((9)/(5sqrt(10)))`
B. `sin^(-1)((9)/(5sqrt(10)))`
C. `tan^(-1)((9)/(5sqrt(10)))`
D. `tan^(-1)(9/14)`

Explanation

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

### Step 1: Convert \( \alpha \) to a tangent form
Given \( \alpha = \cos^{-1}\left(\frac{3}{5}\right) \), we can find the sine of \( \alpha \) using the identity:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
Substituting \( \cos \alpha = \frac{3}{5} \):
\[ \sin^2 \alpha + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 \alpha + \frac{9}{25} = 1 \]
\[ \sin^2 \alpha = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \sin \alpha = \frac{4}{5} \]

Now, we can express \( \alpha \) in terms of tangent:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

### Step 2: Write \( \alpha - \beta \) in terms of tangent
Now we have:
\[ \alpha - \beta = \tan^{-1}\left(\frac{4}{3}\right) - \tan^{-1}\left(\frac{1}{3}\right) \]
We can use the formula for the difference of two inverse tangents:
\[ \tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
Let \( a = \frac{4}{3} \) and \( b = \frac{1}{3} \):
\[ \tan(\alpha - \beta) = \tan^{-1}\left(\frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{3} \cdot \frac{1}{3}}\right) \]
Calculating the numerator:
\[ \frac{4}{3} - \frac{1}{3} = \frac{3}{3} = 1 \]

Calculating the denominator:
\[ 1 + \frac{4}{3} \cdot \frac{1}{3} = 1 + \frac{4}{9} = \frac{9}{9} + \frac{4}{9} = \frac{13}{9} \]

So we have:
\[ \tan(\alpha - \beta) = \tan^{-1}\left(\frac{1}{\frac{13}{9}}\right) = \tan^{-1}\left(\frac{9}{13}\right) \]

### Step 3: Conclusion
Thus, we find that:
\[ \alpha - \beta = \tan^{-1}\left(\frac{9}{13}\right) \]