`tan^(-1) ""1/4 + tan^(-1)"" 2/9 = 1/2 cos^(-1)"" 3/5`

Answer

To solve the equation \( \tan^{-1} \left( \frac{1}{4} \right) + \tan^{-1} \left( \frac{2}{9} \right) = \frac{1}{2} \cos^{-1} \left( \frac{3}{5} \right) \), we will use the formula for the sum of inverse tangents.

### Step 1: Use the formula for the sum of inverse tangents
The formula for the sum of two inverse tangents is:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \]
Here, let \( x = \frac{1}{4} \) and \( y = \frac{2}{9} \).

### Step 2: Calculate \( x + y \) and \( 1 - xy \)
First, we calculate \( x + y \):
\[ x + y = \frac{1}{4} + \frac{2}{9} \]
To add these fractions, we need a common denominator. The least common multiple of 4 and 9 is 36.
\[ x + y = \frac{9}{36} + \frac{8}{36} = \frac{17}{36} \]

Next, we calculate \( xy \):
\[ xy = \frac{1}{4} \cdot \frac{2}{9} = \frac{2}{36} = \frac{1}{18} \]
Now, we find \( 1 - xy \):
\[ 1 - xy = 1 - \frac{1}{18} = \frac{18}{18} - \frac{1}{18} = \frac{17}{18} \]

### Step 3: Substitute into the formula
Now we can substitute \( x + y \) and \( 1 - xy \) back into the formula:
\[ \tan^{-1} \left( \frac{x + y}{1 - xy} \right) = \tan^{-1} \left( \frac{\frac{17}{36}}{\frac{17}{18}} \right) \]
Simplifying this gives:
\[ \tan^{-1} \left( \frac{17}{36} \cdot \frac{18}{17} \right) = \tan^{-1} \left( \frac{18}{36} \right) = \tan^{-1} \left( \frac{1}{2} \right) \]

### Step 4: Equate to the right-hand side
Now we have:
\[ \tan^{-1} \left( \frac{1}{2} \right) = \frac{1}{2} \cos^{-1} \left( \frac{3}{5} \right) \]

### Step 5: Use the double angle formula for cosine
We can use the identity:
\[ 2 \tan^{-1}(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]
Setting \( x = \frac{1}{2} \):
\[ 2 \tan^{-1} \left( \frac{1}{2} \right) = \cos^{-1} \left( \frac{1 - \left( \frac{1}{2} \right)^2}{1 + \left( \frac{1}{2} \right)^2} \right) \]
Calculating this gives:
\[ = \cos^{-1} \left( \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} \right) = \cos^{-1} \left( \frac{\frac{3}{4}}{\frac{5}{4}} \right) = \cos^{-1} \left( \frac{3}{5} \right) \]

### Conclusion
Thus, we have shown that:
\[ \tan^{-1} \left( \frac{1}{4} \right) + \tan^{-1} \left( \frac{2}{9} \right) = \frac{1}{2} \cos^{-1} \left( \frac{3}{5} \right) \]
This verifies the original equation.