`sin^(-1)((8)/(17))+sin^(-1)((3)/(5))=`

Options

A. `tan^(-1)((77)/(36))`
B. `tan^(-1)((77)/(85))`
C. `pi+tan^(-1)((77)/(36))`
D. `pi-tan^(-1)((77)/(36))`

Answer

To solve the equation \( \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) \), we will use the formula for the sum of inverse sine functions.

### Step-by-Step Solution:

1. **Identify the values**: Let \( x = \frac{8}{17} \) and \( y = \frac{3}{5} \).

2. **Check the domain**: Both \( x \) and \( y \) are in the range \([-1, 1]\), so we can apply the formula.

3. **Use the formula**: The formula for \( \sin^{-1}(x) + \sin^{-1}(y) \) is given by:
\[ \sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1\left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right)} \]

4. **Calculate \( \sqrt{1 - y^2} \)**:
\[ \sqrt{1 - y^2} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

5. **Calculate \( \sqrt{1 - x^2} \)**:
\[ \sqrt{1 - x^2} = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17} \]

6. **Substitute into the formula**:
\[ \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1\left(\frac{8}{17} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{15}{17}\right)} \]

7. **Calculate the expression**:
\[ \frac{8}{17} \cdot \frac{4}{5} = \frac{32}{85} \]
\[ \frac{3}{5} \cdot \frac{15}{17} = \frac{45}{85} \]
\[ \frac{32}{85} + \frac{45}{85} = \frac{77}{85} \]

8. **Final result**:
\[ \sin^{-1}\left(\frac{77}{85}\right) \]

9. **Convert to tangent**:
To express this in terms of tangent, we can use the relationship:
\[ \tan(y) = \frac{\text{opposite}}{\text{adjacent}} = \frac{77}{36} \]
where the adjacent side can be calculated as \( \sqrt{85^2 - 77^2} = 36 \).

10. **Conclusion**:
Therefore, we have:
\[ \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{77}{36}\right) \]

### Final Answer:
\[ \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{77}{36}\right) \]