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

Options

A. `pm 1`
B. `pm 2`
C. `pm sqrt3`
D. `pm 1/sqrt2`

Answer

To solve the equation

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

we will 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:

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

### Step 2: Apply the formula

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

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

### Step 3: Simplify the numerator

The numerator becomes:

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

Calculating the numerator:

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

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

### Step 4: Simplify the denominator

Now, we simplify the denominator:

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

Calculating this gives:

\[ = \frac{(x-2)(x+2) - (x^2 - 1)}{(x-2)(x+2)} = \frac{x^2 - 4 - x^2 + 1}{(x-2)(x+2)} = \frac{-3}{(x-2)(x+2)} \]

### Step 5: Combine the results

Now we can combine the results:

\[ \tan^{-1}\left(\frac{\frac{2x^2 - 4}{(x-2)(x+2)}}{\frac{-3}{(x-2)(x+2)}}\right) = \tan^{-1}\left(\frac{2x^2 - 4}{-3}\right) \]

Setting this equal to \(\frac{\pi}{4}\):

\[ \frac{2x^2 - 4}{-3} = 1 \]

### Step 6: Solve for \(x\)

Multiplying both sides by -3 gives:

\[ 2x^2 - 4 = -3 \]

\[ 2x^2 = 1 \quad \Rightarrow \quad x^2 = \frac{1}{2} \]

Taking the square root gives:

\[ x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

### Final Answer

Thus, the solution for \(x\) is:

\[ x = \frac{1}{\sqrt{2}} \text{ or } x = -\frac{1}{\sqrt{2}} \]