`tan^(-1)((1)/(4))+tan^(-1)((2)/(9))` is equal to :

Options

A. `(1)/(2) cos^(-1)((3)/(5))`
B. `(1)/(2) sin^(-1)((3)/(5))`
C. `(1)/(2) tan^(-1)((3)/(5))`
D. `tan^(-1)((1)/(2))`

Answer

To solve the expression \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \), we can use the identity for the sum of inverse tangents:

\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \quad \text{if } xy < 1 \]

### Step 1: Identify \(x\) and \(y\)
Let:
- \( x = \frac{1}{4} \)
- \( y = \frac{2}{9} \)

### Step 2: Check the condition \(xy < 1\)
Calculate \(xy\):
\[ xy = \left(\frac{1}{4}\right) \left(\frac{2}{9}\right) = \frac{2}{36} = \frac{1}{18} \]
Since \( \frac{1}{18} < 1 \), we can use the identity.

### Step 3: Apply the identity
Now, apply the identity:
\[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \cdot \frac{2}{9}}\right) \]

### Step 4: Calculate the numerator
Find a common denominator for the numerator:
\[ \frac{1}{4} + \frac{2}{9} = \frac{9}{36} + \frac{8}{36} = \frac{17}{36} \]

### Step 5: Calculate the denominator
Now calculate the denominator:
\[ 1 - \frac{1}{4} \cdot \frac{2}{9} = 1 - \frac{1}{18} = \frac{18}{18} - \frac{1}{18} = \frac{17}{18} \]

### Step 6: Combine the results
Now substitute back into the formula:
\[ \tan^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) = \tan^{-1}\left(\frac{17 \cdot 18}{36 \cdot 17}\right) = \tan^{-1}\left(\frac{18}{36}\right) = \tan^{-1}\left(\frac{1}{2}\right) \]

### Final Result
Thus, we have:
\[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \tan^{-1}\left(\frac{1}{2}\right) \]