If `sin^(-1) ""3/5+sin^(-1)""5/13=sin^(-1)x, " then " x=`

Options

A. `41/56`
B. `56/65`
C. `13/41`
D. none

Answer

To solve the equation \( \sin^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \sin^{-1}(x) \), we can use the formula for the sum of inverse sine functions:

\[ \sin^{-1}(a) + \sin^{-1}(b) = \sin^{-1}\left(a\sqrt{1-b^2} + b\sqrt{1-a^2}\right) \]

where \( a = \frac{3}{5} \) and \( b = \frac{5}{13} \).

### Step 1: Calculate \( \sqrt{1 - b^2} \)

First, we need to calculate \( \sqrt{1 - b^2} \):

\[ b = \frac{5}{13} \implies b^2 = \left(\frac{5}{13}\right)^2 = \frac{25}{169} \]

Thus,

\[ 1 - b^2 = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169} \]

So,

\[ \sqrt{1 - b^2} = \sqrt{\frac{144}{169}} = \frac{12}{13} \]

### Step 2: Calculate \( \sqrt{1 - a^2} \)

Next, we calculate \( \sqrt{1 - a^2} \):

\[ a = \frac{3}{5} \implies a^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \]

Thus,

\[ 1 - a^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25} \]

So,

\[ \sqrt{1 - a^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 3: Substitute into the formula

Now we substitute \( a \), \( b \), \( \sqrt{1 - b^2} \), and \( \sqrt{1 - a^2} \) into the formula:

\[ x = a\sqrt{1 - b^2} + b\sqrt{1 - a^2} \]

Substituting the values:

\[ x = \frac{3}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{4}{5} \]

Calculating each term:

1. For the first term:

\[ \frac{3 \cdot 12}{5 \cdot 13} = \frac{36}{65} \]

2. For the second term:

\[ \frac{5 \cdot 4}{13 \cdot 5} = \frac{20}{65} \]

### Step 4: Combine the terms

Now, we combine the two terms:

\[ x = \frac{36}{65} + \frac{20}{65} = \frac{56}{65} \]

### Final Answer

Thus, the value of \( x \) is:

\[ \boxed{\frac{56}{65}} \]