Prove that `sin^(- 1)(8/17)+sin^(- 1)(3/5)=sin^(- 1)(77/85)`

Explanation

To prove that \( \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{77}{85}\right) \), we will use the properties of triangles and the tangent addition formula.

### Step 1: Define the angles
Let:
- \( \theta = \sin^{-1}\left(\frac{8}{17}\right) \)
- \( \phi = \sin^{-1}\left(\frac{3}{5}\right) \)

### Step 2: Draw the triangles for \( \theta \) and \( \phi \)
For \( \theta \):
- The opposite side (perpendicular) = 8
- The hypotenuse = 17
- Using the Pythagorean theorem, the base (adjacent side) can be calculated as:
\[ \text{Base} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \]
Thus, we have a right triangle with:
- Opposite = 8
- Base = 15
- Hypotenuse = 17

For \( \phi \):
- The opposite side (perpendicular) = 3
- The hypotenuse = 5
- Using the Pythagorean theorem, the base can be calculated as:
\[ \text{Base} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \]
Thus, we have a right triangle with:
- Opposite = 3
- Base = 4
- Hypotenuse = 5

### Step 3: Convert to tangent functions
Now we can express \( \tan(\theta) \) and \( \tan(\phi) \):
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Base}} = \frac{8}{15} \]
\[ \tan(\phi) = \frac{\text{Opposite}}{\text{Base}} = \frac{3}{4} \]

### Step 4: Use the tangent addition formula
We will use the formula for \( \tan(A + B) \):
\[ \tan(\theta + \phi) = \frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta) \tan(\phi)} \]
Substituting the values:
\[ \tan(\theta + \phi) = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \left(\frac{8}{15} \cdot \frac{3}{4}\right)} \]

### Step 5: Calculate the numerator
To add the fractions in the numerator:
\[ \frac{8}{15} + \frac{3}{4} = \frac{32}{60} + \frac{45}{60} = \frac{77}{60} \]

### Step 6: Calculate the denominator
Now calculate the denominator:
\[ 1 - \left(\frac{8}{15} \cdot \frac{3}{4}\right) = 1 - \frac{24}{60} = \frac{36}{60} \]

### Step 7: Combine the results
Now substituting back into the tangent formula:
\[ \tan(\theta + \phi) = \frac{\frac{77}{60}}{\frac{36}{60}} = \frac{77}{36} \]

### Step 8: Find \( \sin(\theta + \phi) \)
Now we need to find \( \sin(\theta + \phi) \):
Using the identity \( \sin(x) = \frac{\text{Opposite}}{\text{Hypotenuse}} \):
- We have \( \tan(\theta + \phi) = \frac{77}{36} \).
- To find the hypotenuse, we can use the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{77^2 + 36^2} = \sqrt{5929 + 1296} = \sqrt{7225} = 85 \]

### Step 9: Final calculation
Thus, we have:
\[ \sin(\theta + \phi) = \frac{77}{85} \]

### Conclusion
Therefore, we can conclude:
\[ \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{77}{85}\right) \]
This completes the proof.