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 can follow these steps:

### Step 1: Rewrite the expression
We can rewrite \( 2\tan^{-1}\left(\frac{1}{2}\right) \) as:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{2}\right) \]
Thus, the expression becomes:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{4}\right) \]

### Step 2: Apply the formula for the sum of arctangents
We use the formula:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
for \( x = \frac{1}{2} \) and \( y = \frac{1}{2} \).

### Step 3: Calculate \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{2}\right) \)
Using the formula:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{2}}{1 - \frac{1}{2} \cdot \frac{1}{2}}\right) \]
Calculating the numerator:
\[ \frac{1}{2} + \frac{1}{2} = 1 \]
Calculating the denominator:
\[ 1 - \frac{1}{4} = \frac{3}{4} \]
Thus, we have:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{1}{\frac{3}{4}}\right) = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 4: Add \( \tan^{-1}\left(\frac{4}{3}\right) \) and \( \tan^{-1}\left(\frac{1}{4}\right) \)
Now we need to evaluate:
\[ \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{4}\right) \]
Using the formula again:
\[ \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{4}\right) = \tan^{-1}\left(\frac{\frac{4}{3} + \frac{1}{4}}{1 - \frac{4}{3} \cdot \frac{1}{4}}\right) \]

### Step 5: Calculate the numerator and denominator
Calculating the numerator:
\[ \frac{4}{3} + \frac{1}{4} = \frac{16 + 3}{12} = \frac{19}{12} \]
Calculating the denominator:
\[ 1 - \frac{4}{12} = 1 - \frac{1}{3} = \frac{2}{3} \]
Thus, we have:
\[ \tan^{-1}\left(\frac{\frac{19}{12}}{\frac{2}{3}}\right) = \tan^{-1}\left(\frac{19 \cdot 3}{12 \cdot 2}\right) = \tan^{-1}\left(\frac{57}{24}\right) \]

### Final Result
So, the final result of the expression \( 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{4}\right) \) is:
\[ \tan^{-1}\left(\frac{57}{24}\right) \]