Solve \( \tan^{-1}\left(\frac{1}{4}\right) + 2\tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{6}\right) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{4} \)

Solution:

Step 1: Evaluate \(2\tan^{-1}(1/5)\)

\[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \Rightarrow \tan(2\tan^{-1}(1/5)) = \frac{2/5}{1 – 1/25} = \frac{2/5}{24/25} = \frac{5}{12} \]

\[ \Rightarrow 2\tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{5}{12}\right) \]

Step 2: Combine first two terms

\[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{5}{12}\right) \]

\[ = \tan^{-1}\left(\frac{\frac{1}{4}+\frac{5}{12}}{1 – \frac{5}{48}}\right) = \tan^{-1}\left(\frac{2/3}{43/48}\right) = \tan^{-1}\left(\frac{32}{43}\right) \]

Step 3: Add next term

\[ \tan^{-1}\left(\frac{32}{43}\right) + \tan^{-1}\left(\frac{1}{6}\right) \]

\[ = \tan^{-1}\left(\frac{\frac{32}{43}+\frac{1}{6}}{1 – \frac{32}{258}}\right) = \tan^{-1}\left(\frac{235/258}{113/129}\right) = \tan^{-1}\left(\frac{235}{226}\right) \]

Step 4: Add last term

\[ \tan^{-1}\left(\frac{235}{226}\right) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{4} \]

Using identity:

\[ \tan^{-1}a + \tan^{-1}b = \frac{\pi}{4} \Rightarrow \frac{a+b}{1 – ab} = 1 \]

\[ \frac{\frac{235}{226} + \frac{1}{x}}{1 – \frac{235}{226x}} = 1 \]

Solve:

\[ \frac{235x + 226}{226x – 235} = 1 \]

\[ 235x + 226 = 226x – 235 \]

\[ 9x = -461 \Rightarrow x = -\frac{461}{9} \]

Final Answer:

\[ x = -\frac{461}{9} \]