Find the value of `sin^(-1) ((3)/(5)) + tan^(-1) ((1)/(7))`

Explanation

To find the value of \( \sin^{-1} \left( \frac{3}{5} \right) + \tan^{-1} \left( \frac{1}{7} \right) \), we can follow these steps:

### Step 1: Identify the angle for \( \sin^{-1} \left( \frac{3}{5} \right) \)
Let \( \theta = \sin^{-1} \left( \frac{3}{5} \right) \). By definition of the inverse sine function, we have:
\[ \sin \theta = \frac{3}{5} \]
This means that in a right triangle where the opposite side is 3 and the hypotenuse is 5, we can find the adjacent side using the Pythagorean theorem.

### Step 2: Calculate the adjacent side
Using the Pythagorean theorem:
\[ \text{adjacent}^2 + 3^2 = 5^2 \]
\[ \text{adjacent}^2 + 9 = 25 \]
\[ \text{adjacent}^2 = 16 \quad \Rightarrow \quad \text{adjacent} = 4 \]

### Step 3: Find \( \tan \theta \)
Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4} \]

### Step 4: Rewrite the expression
Now we can rewrite the original expression:
\[ \sin^{-1} \left( \frac{3}{5} \right) + \tan^{-1} \left( \frac{1}{7} \right) = \tan^{-1} \left( \frac{3}{4} \right) + \tan^{-1} \left( \frac{1}{7} \right) \]

### Step 5: Use the formula for \( \tan^{-1} a + \tan^{-1} b \)
We can use the formula:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
where \( a = \frac{3}{4} \) and \( b = \frac{1}{7} \).

### Step 6: Substitute values into the formula
Substituting \( a \) and \( b \):
\[ \tan^{-1} \left( \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \cdot \frac{1}{7}} \right) \]

### Step 7: Calculate the numerator
Calculating the numerator:
\[ \frac{3}{4} + \frac{1}{7} = \frac{21 + 4}{28} = \frac{25}{28} \]

### Step 8: Calculate the denominator
Calculating the denominator:
\[ 1 - \frac{3}{4} \cdot \frac{1}{7} = 1 - \frac{3}{28} = \frac{28 - 3}{28} = \frac{25}{28} \]

### Step 9: Combine the results
Now substituting back into the formula:
\[ \tan^{-1} \left( \frac{\frac{25}{28}}{\frac{25}{28}} \right) = \tan^{-1}(1) \]

### Step 10: Final result
Since \( \tan^{-1}(1) = \frac{\pi}{4} \):
\[ \sin^{-1} \left( \frac{3}{5} \right) + \tan^{-1} \left( \frac{1}{7} \right) = \frac{\pi}{4} \]