Evaluate `2tan^(-1)(1/2)+tan^(-1)(1/4)`

Answer

To evaluate the expression \( 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{4}\right) \), we will follow these steps:

### Step 1: Use the double angle formula for tangent inverse
We can use the formula for \( 2\tan^{-1}(x) \):
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]
In our case, \( x = \frac{1}{2} \).

### Step 2: Substitute \( x \) into the formula
Substituting \( x = \frac{1}{2} \) into the formula gives:
\[ 2\tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2}\right) \]

### Step 3: Simplify the expression
Calculating the numerator and denominator:
- Numerator: \( 2 \cdot \frac{1}{2} = 1 \)
- Denominator: \( 1 - \frac{1}{4} = \frac{3}{4} \)

Thus, we have:
\[ 2\tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{1}{\frac{3}{4}}\right) = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 4: Combine with the other term
Now, we substitute back into the original expression:
\[ 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{4}\right) = \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{4}\right) \]

### Step 5: Use the addition formula for tangent inverse
We can use the formula for the sum of two inverse tangents:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
Here, \( x = \frac{4}{3} \) and \( y = \frac{1}{4} \).

### Step 6: Substitute into the addition formula
Substituting \( x \) and \( y \):
\[ \tan^{-1}\left(\frac{\frac{4}{3} + \frac{1}{4}}{1 - \frac{4}{3} \cdot \frac{1}{4}}\right) \]

### Step 7: Simplify the numerator and denominator
Calculating the numerator:
\[ \frac{4}{3} + \frac{1}{4} = \frac{16 + 3}{12} = \frac{19}{12} \]
Calculating the denominator:
\[ 1 - \frac{4}{3} \cdot \frac{1}{4} = 1 - 1 = 0 \]
This indicates that we need to be careful with the calculation. Let's recalculate the denominator correctly:
\[ 1 - \frac{4}{12} = \frac{3}{12} = \frac{1}{4} \]

### Step 8: Final expression
Now we substitute back:
\[ \tan^{-1}\left(\frac{\frac{19}{12}}{\frac{1}{4}}\right) = \tan^{-1}\left(\frac{19 \cdot 4}{12}\right) = \tan^{-1}\left(\frac{76}{12}\right) = \tan^{-1}\left(\frac{19}{3}\right) \]

Thus, the final answer is:
\[ \tan^{-1}\left(\frac{19}{3}\right) \]