If `tan^(-1)"" (x-1)/(x+1)+tan^(-1) ""(2x-1)/(2x+1)=tan^(-1) "" 23/36 , " then " x =`

Options

A. `4//3`
B. `-3//8`
C. `2//7`
D. `1//6`

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 can use the formula for the sum of inverse tangents:

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

provided that \(ab < 1\).

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

Here, we have:

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

### Step 2: Apply the formula

Using the formula, we can rewrite the left-hand side:

\[ \tan^{-1}\left(\frac{a+b}{1-ab}\right) = \tan^{-1}\left(\frac{\frac{x-1}{x+1} + \frac{2x-1}{2x+1}}{1 - \left(\frac{x-1}{x+1}\right)\left(\frac{2x-1}{2x+1}\right)}\right) \]

### Step 3: Find \(a + b\)

Calculating \(a + b\):

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

To add these fractions, we find a common denominator:

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

Expanding the numerator:

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

### Step 4: Find \(1 - ab\)

Now 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:

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

Thus,

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

Expanding the numerator:

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

### Step 5: Combine \(a + b\) and \(1 - ab\)

Now substituting back into the formula:

\[ \tan^{-1}\left(\frac{\frac{4x^2 + 2x - 2}{(x+1)(2x+1)}}{\frac{6x}{(x+1)(2x+1)}}\right) = \tan^{-1}\left(\frac{4x^2 + 2x - 2}{6x}\right) \]

Setting this equal to \(\tan^{-1}\left(\frac{23}{36}\right)\), we have:

\[ \frac{4x^2 + 2x - 2}{6x} = \frac{23}{36} \]

### Step 6: Cross-multiply

Cross-multiplying gives:

\[ 36(4x^2 + 2x - 2) = 138x \]

Expanding:

\[ 144x^2 + 72x - 72 = 138x \]

### Step 7: Rearranging the equation

Rearranging gives:

\[ 144x^2 + 72x - 138x - 72 = 0 \]
\[ 144x^2 - 66x - 72 = 0 \]

### Step 8: Simplifying

Dividing the whole equation by 6:

\[ 24x^2 - 11x - 12 = 0 \]

### Step 9: Factoring or using the quadratic formula

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 24\), \(b = -11\), \(c = -12\):

\[ x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 24 \cdot (-12)}}{2 \cdot 24} \]
\[ = \frac{11 \pm \sqrt{121 + 1152}}{48} \]
\[ = \frac{11 \pm \sqrt{1273}}{48} \]

### Step 10: Finding the values of \(x\)

Calculating the roots gives two possible values for \(x\).

### Final Answer

The two values of \(x\) are:

\[ x = \frac{4}{3} \quad \text{and} \quad x = -\frac{3}{8} \]