If `tan^(-1) ""1/4 + 2 tan^(-1)"" 1/5+tan^(-1)"" 1/6 + tan^(-1) ""1/x=pi/4 , " then " x=`

Options

A. `- 321/9`
B. `237/4`
C. `- 461/9`
D. `349/7`

Answer

To solve the equation \( \tan^{-1} \left( \frac{1}{4} \right) + 2 \tan^{-1} \left( \frac{1}{5} \right) + \tan^{-1} \left( \frac{1}{6} \right) + \tan^{-1} \left( \frac{1}{x} \right) = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Simplify \( 2 \tan^{-1} \left( \frac{1}{5} \right) \)
Using the double angle formula for tangent:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Let \( \theta = \tan^{-1} \left( \frac{1}{5} \right) \). Then, \( \tan(\theta) = \frac{1}{5} \).

Calculating \( \tan(2\theta) \):
\[ \tan(2\tan^{-1} \left( \frac{1}{5} \right)) = \frac{2 \cdot \frac{1}{5}}{1 - \left( \frac{1}{5} \right)^2} = \frac{\frac{2}{5}}{1 - \frac{1}{25}} = \frac{\frac{2}{5}}{\frac{24}{25}} = \frac{2 \cdot 25}{5 \cdot 24} = \frac{10}{24} = \frac{5}{12} \]

### Step 2: Substitute back into the equation
Now, substitute \( 2 \tan^{-1} \left( \frac{1}{5} \right) \) with \( \tan^{-1} \left( \frac{5}{12} \right) \):
\[ \tan^{-1} \left( \frac{1}{4} \right) + \tan^{-1} \left( \frac{5}{12} \right) + \tan^{-1} \left( \frac{1}{6} \right) + \tan^{-1} \left( \frac{1}{x} \right) = \frac{\pi}{4} \]

### Step 3: Combine the first two terms
Using the formula for the sum of arctangents:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
Let \( a = \frac{1}{4} \) and \( b = \frac{5}{12} \):
\[ \tan^{-1} \left( \frac{\frac{1}{4} + \frac{5}{12}}{1 - \frac{1}{4} \cdot \frac{5}{12}} \right) \]
Calculating \( \frac{1}{4} + \frac{5}{12} \):
\[ \frac{3}{12} + \frac{5}{12} = \frac{8}{12} = \frac{2}{3} \]
Calculating \( 1 - \frac{1}{4} \cdot \frac{5}{12} \):
\[ 1 - \frac{5}{48} = \frac{48 - 5}{48} = \frac{43}{48} \]
Thus,
\[ \tan^{-1} \left( \frac{\frac{2}{3}}{\frac{43}{48}} \right) = \tan^{-1} \left( \frac{2 \cdot 48}{3 \cdot 43} \right) = \tan^{-1} \left( \frac{96}{129} \right) = \tan^{-1} \left( \frac{32}{43} \right) \]

### Step 4: Combine with the third term
Now we have:
\[ \tan^{-1} \left( \frac{32}{43} \right) + \tan^{-1} \left( \frac{1}{6} \right) + \tan^{-1} \left( \frac{1}{x} \right) = \frac{\pi}{4} \]
Using the sum formula again:
Let \( a = \frac{32}{43} \) and \( b = \frac{1}{6} \):
\[ \tan^{-1} \left( \frac{\frac{32}{43} + \frac{1}{6}}{1 - \frac{32}{43} \cdot \frac{1}{6}} \right) \]
Calculating \( \frac{32}{43} + \frac{1}{6} \):
\[ \frac{32 \cdot 6 + 1 \cdot 43}{43 \cdot 6} = \frac{192 + 43}{258} = \frac{235}{258} \]
Calculating \( 1 - \frac{32}{43} \cdot \frac{1}{6} \):
\[ 1 - \frac{32}{258} = \frac{258 - 32}{258} = \frac{226}{258} \]
Thus,
\[ \tan^{-1} \left( \frac{\frac{235}{258}}{\frac{226}{258}} \right) = \tan^{-1} \left( \frac{235}{226} \right) \]

### Step 5: Set up the final equation
Now we have:
\[ \tan^{-1} \left( \frac{235}{226} \right) + \tan^{-1} \left( \frac{1}{x} \right) = \frac{\pi}{4} \]
Using the property:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \frac{\pi}{4} \implies ab = 1 \]
Thus,
\[ \frac{235}{226} \cdot \frac{1}{x} = 1 \implies x = \frac{235}{226} \]

### Final Answer
\[ x = \frac{235}{226} \]