`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 will follow these steps:

### Step 1: Set the angle for sine inverse
Let \( \theta = \sin^{-1}\left(\frac{3}{5}\right) \). This implies that:
\[ \sin \theta = \frac{3}{5} \]

### Step 2: Draw a right triangle
In a right triangle, the sine of an angle is defined as the ratio of the length of the opposite side (perpendicular) to the length of the hypotenuse. Here, we can denote:
- Opposite side = 3
- Hypotenuse = 5

### Step 3: Find the base using Pythagorean theorem
Using the Pythagorean theorem:
\[ \text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Base}^2 \]
Substituting the known values:
\[ 5^2 = 3^2 + \text{Base}^2 \implies 25 = 9 + \text{Base}^2 \]
\[ \text{Base}^2 = 25 - 9 = 16 \implies \text{Base} = 4 \]

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

### Step 5: Rewrite the expression
Since \( \theta = \sin^{-1}\left(\frac{3}{5}\right) \), we can rewrite the 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 for the sum of two arctangents:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
Let \( x = \frac{3}{4} \) and \( y = \frac{1}{7} \):
\[ \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \cdot \frac{1}{7}}\right) \]

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

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

### Step 8: Combine the results
Now substituting back into the arctangent formula:
\[ \tan^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

### Final Answer
Thus, the value of \( \sin^{-1}\left(\frac{3}{5}\right) + \tan^{-1}\left(\frac{1}{7}\right) \) is:
\[ \frac{\pi}{4} \]