`Sin^(-1)((12)/(13))-sin^(-1)((3)/(5))` is equal to

Options

A. `(pi)/(2)-cos^(-1)((33)/(65))`
B. `pi-cos^(-1)((33)/(65))`
C. `(pi)/(2)-sin^(-1)((33)/(65))`
D. `(pi)/(2)+sin^(-1)((63)/(65))`

Answer

To solve the expression \( \sin^{-1}\left(\frac{12}{13}\right) - \sin^{-1}\left(\frac{3}{5}\right) \), we can use the formula for the difference of inverse sine functions:

\[ \sin^{-1}(x) - \sin^{-1}(y) = \sin^{-1}\left(\sqrt{x^2 + y^2 - 2xy}\right) \]

where \( x = \frac{12}{13} \) and \( y = \frac{3}{5} \).

### Step 1: Identify \( x \) and \( y \)
Let:
- \( x = \frac{12}{13} \)
- \( y = \frac{3}{5} \)

### Step 2: Calculate \( x^2 \) and \( y^2 \)
\[ x^2 = \left(\frac{12}{13}\right)^2 = \frac{144}{169} \]
\[ y^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \]

### Step 3: Find a common denominator for \( y^2 \)
The common denominator for \( 169 \) and \( 25 \) is \( 4225 \).
Convert \( y^2 \):
\[ y^2 = \frac{9}{25} = \frac{9 \times 169}{25 \times 169} = \frac{1521}{4225} \]

### Step 4: Calculate \( 2xy \)
\[ 2xy = 2 \cdot \frac{12}{13} \cdot \frac{3}{5} = \frac{72}{65} \]

Convert \( \frac{72}{65} \) to have a common denominator of \( 4225 \):
\[ 2xy = \frac{72 \times 65}{65 \times 65} = \frac{4680}{4225} \]

### Step 5: Combine the terms
Now we can combine \( x^2 + y^2 - 2xy \):
\[ x^2 + y^2 - 2xy = \frac{144}{169} + \frac{9}{25} - \frac{72}{65} \]
Convert \( x^2 \) and \( y^2 \) to have the common denominator \( 4225 \):
\[ x^2 = \frac{144 \times 25}{169 \times 25} = \frac{3600}{4225} \]
\[ y^2 = \frac{9 \times 169}{25 \times 169} = \frac{1521}{4225} \]
Now, we have:
\[ x^2 + y^2 = \frac{3600 + 1521}{4225} = \frac{5121}{4225} \]
Now, subtract \( 2xy \):
\[ x^2 + y^2 - 2xy = \frac{5121}{4225} - \frac{4680}{4225} = \frac{441}{4225} \]

### Step 6: Take the square root
Now we take the square root:
\[ \sqrt{x^2 + y^2 - 2xy} = \sqrt{\frac{441}{4225}} = \frac{21}{65} \]

### Step 7: Final result
Thus, we have:
\[ \sin^{-1}\left(\frac{12}{13}\right) - \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{21}{65}\right) \]