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

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `0`

Answer

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

### Step 1: Combine the first two terms using the formula for the sum of inverse tangents

The formula for the sum of two inverse tangents is:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \quad \text{if } ab < 1 \]

Let \( a = \frac{3}{4} \) and \( b = \frac{3}{5} \).

First, we calculate \( ab \):
\[ ab = \left(\frac{3}{4}\right) \left(\frac{3}{5}\right) = \frac{9}{20} < 1 \]

Now we can apply the formula:
\[ \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{3}{4} \cdot \frac{3}{5}}\right) \]

### Step 2: Calculate the numerator and denominator

**Numerator:**
\[ \frac{3}{4} + \frac{3}{5} = \frac{15}{20} + \frac{12}{20} = \frac{27}{20} \]

**Denominator:**
\[ 1 - \frac{9}{20} = \frac{20}{20} - \frac{9}{20} = \frac{11}{20} \]

### Step 3: Substitute back into the formula

Now substituting the values back:
\[ \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{\frac{27}{20}}{\frac{11}{20}}\right) = \tan^{-1}\left(\frac{27}{11}\right) \]

### Step 4: Now subtract the third term

We need to calculate:
\[ \tan^{-1}\left(\frac{27}{11}\right) - \tan^{-1}\left(\frac{8}{19}\right) \]

Using the formula for the difference of two inverse tangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \quad \text{if } ab > -1 \]

Let \( a = \frac{27}{11} \) and \( b = \frac{8}{19} \).

### Step 5: Calculate \( ab \)

\[ ab = \left(\frac{27}{11}\right) \left(\frac{8}{19}\right) = \frac{216}{209} > -1 \]

Now we can apply the difference formula:
\[ \tan^{-1}\left(\frac{27}{11}\right) - \tan^{-1}\left(\frac{8}{19}\right) = \tan^{-1}\left(\frac{\frac{27}{11} - \frac{8}{19}}{1 + \frac{27}{11} \cdot \frac{8}{19}}\right) \]

### Step 6: Calculate the numerator and denominator for the difference

**Numerator:**
\[ \frac{27}{11} - \frac{8}{19} = \frac{27 \cdot 19 - 8 \cdot 11}{11 \cdot 19} = \frac{513 - 88}{209} = \frac{425}{209} \]

**Denominator:**
\[ 1 + \frac{216}{209} = \frac{209 + 216}{209} = \frac{425}{209} \]

### Step 7: Substitute back into the formula

Now substituting back:
\[ \tan^{-1}\left(\frac{27}{11}\right) - \tan^{-1}\left(\frac{8}{19}\right) = \tan^{-1}\left(\frac{\frac{425}{209}}{\frac{425}{209}}\right) = \tan^{-1}(1) \]

### Step 8: Final answer

Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we conclude:
\[ \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{3}{5}\right) - \tan^{-1}\left(\frac{8}{19}\right) = \frac{\pi}{4} \]

### Final Answer:
\[ \frac{\pi}{4} \]