IF ` tan ^(-1) 2x + tan ^(-1) 3x =(pi)/(4) , ` then x=

Options

A. `-1`
B. `(1)/(3)`
C. `(1)/(6)`
D. `(1)/(2)`

Answer

To solve the equation \( \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \), we can follow these steps:

### Step 1: Use the tangent addition formula
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
if \( ab < 1 \). Here, let \( a = 2x \) and \( b = 3x \).

### Step 2: Apply the formula
Using the formula, we can rewrite the left-hand side:
\[ \tan^{-1}(2x) + \tan^{-1}(3x) = \tan^{-1}\left(\frac{2x + 3x}{1 - 2x \cdot 3x}\right) = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]

### Step 3: Set the equation
Now, we can set this equal to \( \frac{\pi}{4} \):
\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) = \frac{\pi}{4} \]

### Step 4: Take the tangent of both sides
Taking the tangent of both sides, we get:
\[ \frac{5x}{1 - 6x^2} = 1 \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ 5x = 1 - 6x^2 \]

### Step 6: Rearrange the equation
Rearranging this equation results in:
\[ 6x^2 + 5x - 1 = 0 \]

### Step 7: Solve the quadratic equation
Now we can solve the quadratic equation \( 6x^2 + 5x - 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6 \), \( b = 5 \), and \( c = -1 \).

### Step 8: Calculate the discriminant
First, calculate the discriminant:
\[ D = b^2 - 4ac = 5^2 - 4 \cdot 6 \cdot (-1) = 25 + 24 = 49 \]

### Step 9: Find the roots
Now substituting into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{49}}{2 \cdot 6} = \frac{-5 \pm 7}{12} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{2}{12} = \frac{1}{6} \)
2. \( x = \frac{-12}{12} = -1 \)

### Final Answer
Thus, the solutions for \( x \) are:
\[ x = -1 \quad \text{and} \quad x = \frac{1}{6} \]