`sin^(-1)((5)/(13))+cos^(-1)((3)/(5))=`

Options

A. `tan^(-1)((5)/(12))`
B. `tan^(-1)((63)/(16))`
C. `tan^(-1)((4)/(3))`
D. `tan^(-1)((9)/(4))`

Answer

To solve the expression \( \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) \), we can follow these steps:

### Step 1: Let \( y = \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) \)

This is our starting point where we define \( y \) as the sum of the two inverse trigonometric functions.

**Hint:** Define the sum as a variable to simplify the expression.

### Step 2: Let \( \alpha = \sin^{-1}\left(\frac{5}{13}\right) \) and \( \beta = \cos^{-1}\left(\frac{3}{5}\right) \)

Now, we can express \( y \) in terms of \( \alpha \) and \( \beta \):
\[ y = \alpha + \beta \]

**Hint:** Breaking down the components helps in visualizing the problem.

### Step 3: Find the values of \( \sin \alpha \) and \( \cos \beta \)

From the definitions:
- \( \sin \alpha = \frac{5}{13} \)
- \( \cos \beta = \frac{3}{5} \)

**Hint:** Recall the definitions of sine and cosine for the angles.

### Step 4: Determine \( \cos \alpha \) and \( \sin \beta \)

Using the Pythagorean identity, we can find:
- For \( \alpha \):
\[ \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \]

- For \( \beta \):
\[ \sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

**Hint:** Use the Pythagorean theorem to find the missing sine or cosine values.

### Step 5: Use the tangent addition formula

Now, we can find \( \tan \alpha \) and \( \tan \beta \):
- \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \)
- \( \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \)

**Hint:** The tangent of an angle can be found by dividing sine by cosine.

### Step 6: Apply the tangent addition formula

Using the formula \( \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \):
\[ \tan(y) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \left(\frac{5}{12} \cdot \frac{4}{3}\right)} \]

### Step 7: Simplify the expression

First, find a common denominator for the numerator:
\[ \frac{5}{12} + \frac{4}{3} = \frac{5}{12} + \frac{16}{12} = \frac{21}{12} \]

Now, calculate the denominator:
\[ 1 - \left(\frac{5 \cdot 4}{12 \cdot 3}\right) = 1 - \frac{20}{36} = 1 - \frac{5}{9} = \frac{4}{9} \]

Now substituting back:
\[ \tan(y) = \frac{\frac{21}{12}}{\frac{4}{9}} = \frac{21 \cdot 9}{12 \cdot 4} = \frac{189}{48} = \frac{63}{16} \]

### Step 8: Conclusion

Thus, we have:
\[ y = \tan^{-1}\left(\frac{63}{16}\right) \]

**Final Answer:**
\[ \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right) \]