`sin^(-1)(3/5)+tan^(-1)(1/7)=`

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `cos^(-1)(4/5)`
D. `pi`

Answer

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

### Step 1: Let \( \theta = \sin^{-1} \left( \frac{3}{5} \right) \)
This means that \( \sin \theta = \frac{3}{5} \).

### Step 2: Construct a right triangle
In a right triangle, if \( \theta \) is one of the angles, the opposite side (perpendicular) can be taken as 3 and the hypotenuse as 5.

### Step 3: Find the length of the base
Using the Pythagorean theorem:
\[ h^2 = p^2 + b^2 \]
where \( h = 5 \) (hypotenuse) and \( p = 3 \) (opposite side). We need to find the base \( b \):
\[ 5^2 = 3^2 + b^2 \\ 25 = 9 + b^2 \\ b^2 = 25 - 9 = 16 \\ b = 4 \]

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

### Step 5: 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 6: Use the formula for the sum of arctangents
Using the formula:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
where \( a = \frac{3}{4} \) and \( b = \frac{1}{7} \):
\[ \tan^{-1} \left( \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \cdot \frac{1}{7}} \right) \]

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

### Step 8: Calculate the denominator
Calculate 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 we have:
\[ \tan^{-1} \left( \frac{\frac{25}{28}}{\frac{25}{28}} \right) = \tan^{-1}(1) \]

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