`tan^(-1) ""3/4 + tan^(-1) ""3/5 - tan^(-1) ""8/19=`

Options

A. `pi/2`
B. `(3pi)/8`
C. `(5pi)/8`
D. `pi/4`

Answer

To solve the expression \( \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{3}{5} - \tan^{-1} \frac{8}{19} \), we can use the properties of inverse tangent functions.

### Step 1: Use the addition formula for inverse tangent
We start by applying the formula for the sum of two inverse tangents:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \]
for \( xy < 1 \).

Let \( x = \frac{3}{4} \) and \( y = \frac{3}{5} \).

### Step 2: Calculate \( x + y \) and \( 1 - xy \)
Calculate \( x + y \):
\[ x + y = \frac{3}{4} + \frac{3}{5} \]
To add these fractions, find a common denominator:
\[ \text{LCM of } 4 \text{ and } 5 = 20 \]
Thus,
\[ x + y = \frac{3 \times 5}{20} + \frac{3 \times 4}{20} = \frac{15 + 12}{20} = \frac{27}{20} \]

Now calculate \( xy \):
\[ xy = \frac{3}{4} \cdot \frac{3}{5} = \frac{9}{20} \]
Now calculate \( 1 - xy \):
\[ 1 - xy = 1 - \frac{9}{20} = \frac{20 - 9}{20} = \frac{11}{20} \]

### Step 3: Substitute into the formula
Now substitute \( x + y \) and \( 1 - xy \) into the formula:
\[ \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{3}{5} = \tan^{-1} \left( \frac{\frac{27}{20}}{\frac{11}{20}} \right) = \tan^{-1} \left( \frac{27}{11} \right) \]

### Step 4: Combine with the third term
Now we need to subtract \( \tan^{-1} \frac{8}{19} \):
\[ \tan^{-1} \frac{27}{11} - \tan^{-1} \frac{8}{19} \]
We use the subtraction formula for inverse tangent:
\[ \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \]

Let \( x = \frac{27}{11} \) and \( y = \frac{8}{19} \).

### Step 5: Calculate \( x - y \) and \( 1 + xy \)
Calculate \( x - y \):
\[ x - y = \frac{27}{11} - \frac{8}{19} \]
Finding a common denominator:
\[ \text{LCM of } 11 \text{ and } 19 = 209 \]
Thus,
\[ x - y = \frac{27 \times 19}{209} - \frac{8 \times 11}{209} = \frac{513 - 88}{209} = \frac{425}{209} \]

Now calculate \( xy \):
\[ xy = \frac{27}{11} \cdot \frac{8}{19} = \frac{216}{209} \]
Now calculate \( 1 + xy \):
\[ 1 + xy = 1 + \frac{216}{209} = \frac{209 + 216}{209} = \frac{425}{209} \]

### Step 6: Substitute into the formula
Now substitute \( x - y \) and \( 1 + xy \) into the formula:
\[ \tan^{-1} \frac{27}{11} - \tan^{-1} \frac{8}{19} = \tan^{-1} \left( \frac{\frac{425}{209}}{\frac{425}{209}} \right) = \tan^{-1}(1) \]

### Step 7: Final result
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:
\[ \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{3}{5} - \tan^{-1} \frac{8}{19} = \frac{\pi}{4} \]

### Conclusion
The final answer is:
\[ \frac{\pi}{4} \]