If: `tan^(-1)((x-1)/(x+1))+ tan^(-1)((2x-1)/(2x+1)) = tan^(-1) (23/36)`. Then: x=

Options

A. `3/4, -3/8`
B. `3/4, 3/8`
C. `4/3, 3/8`
D. none of these

Answer

To solve the equation

\[ \tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right), \]

we will use the formula for the sum of arctangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right), \]

provided that \(ab < 1\).

### Step 1: Identify \(a\) and \(b\)

Let
\[ a = \frac{x-1}{x+1} \quad \text{and} \quad b = \frac{2x-1}{2x+1}. \]

### Step 2: Calculate \(a + b\)

\[ a + b = \frac{x-1}{x+1} + \frac{2x-1}{2x+1}. \]

To add these fractions, we need a common denominator:

\[ a + b = \frac{(x-1)(2x+1) + (2x-1)(x+1)}{(x+1)(2x+1)}. \]

Expanding the numerators:

\[ (x-1)(2x+1) = 2x^2 + x - 2x - 1 = 2x^2 - x - 1, \]
\[ (2x-1)(x+1) = 2x^2 + 2x - x - 1 = 2x^2 + x - 1. \]

Combining these:

\[ a + b = \frac{(2x^2 - x - 1) + (2x^2 + x - 1)}{(x+1)(2x+1)} = \frac{4x^2 - 2}{(x+1)(2x+1)} = \frac{2(2x^2 - 1)}{(x+1)(2x+1)}. \]

### Step 3: Calculate \(1 - ab\)

Next, we calculate \(ab\):

\[ ab = \left(\frac{x-1}{x+1}\right)\left(\frac{2x-1}{2x+1}\right) = \frac{(x-1)(2x-1)}{(x+1)(2x+1)}. \]

Expanding the numerator:

\[ (x-1)(2x-1) = 2x^2 - x - 2x + 1 = 2x^2 - 3x + 1. \]

Thus,

\[ ab = \frac{2x^2 - 3x + 1}{(x+1)(2x+1)}. \]

Now, we calculate \(1 - ab\):

\[ 1 - ab = 1 - \frac{2x^2 - 3x + 1}{(x+1)(2x+1)} = \frac{(x+1)(2x+1) - (2x^2 - 3x + 1)}{(x+1)(2x+1)}. \]

Expanding the numerator:

\[ (x+1)(2x+1) = 2x^2 + x + 2x + 1 = 2x^2 + 3x + 1. \]

Thus,

\[ 1 - ab = \frac{(2x^2 + 3x + 1) - (2x^2 - 3x + 1)}{(x+1)(2x+1)} = \frac{6x}{(x+1)(2x+1)}. \]

### Step 4: Set up the equation

Now we can substitute \(a + b\) and \(1 - ab\) into the arctangent sum formula:

\[ \tan^{-1}\left(\frac{2(2x^2 - 1)}{(x+1)(2x+1)}\right) = \tan^{-1}\left(\frac{23}{36}\right). \]

This gives us:

\[ \frac{2(2x^2 - 1)}{(x+1)(2x+1)} = \frac{23}{36}. \]

### Step 5: Cross-multiply and simplify

Cross-multiplying gives:

\[ 2(2x^2 - 1) \cdot 36 = 23 \cdot (x+1)(2x+1). \]

Expanding both sides:

\[ 72(2x^2 - 1) = 23(2x^2 + 3x + 1). \]

Distributing:

\[ 144x^2 - 72 = 46x^2 + 69x + 23. \]

### Step 6: Rearranging the equation

Rearranging gives:

\[ 144x^2 - 46x^2 - 69x - 72 - 23 = 0, \]
\[ 98x^2 - 69x - 95 = 0. \]

### Step 7: Solving the quadratic equation

Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{69 \pm \sqrt{(-69)^2 - 4 \cdot 98 \cdot (-95)}}{2 \cdot 98}. \]

Calculating the discriminant:

\[ b^2 - 4ac = 4761 + 37240 = 42001. \]

Thus,

\[ x = \frac{69 \pm \sqrt{42001}}{196}. \]

### Step 8: Finding the roots

Calculating the roots will give us the values of \(x\).