If `tan^(-1)((x+1)/(x-1))+tan^(-1)((x-1)/(x))=tan^(-1)(-7)+pi` then x =

Options

A. 2
B. 3
C. 4
D. none of these

Explanation

To solve the equation

\[ \tan^{-1}\left(\frac{x+1}{x-1}\right) + \tan^{-1}\left(\frac{x-1}{x}\right) = \tan^{-1}(-7) + \pi, \]

we can use the formula for the sum of inverse tangents:

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

where \( n \) is an integer, provided that \( ab < 1 \).

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

Let

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

### Step 2: Check the product \( ab \)

Calculate \( ab \):

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

### Step 3: Apply the sum formula

Using the sum formula, we have:

\[ \tan^{-1}\left(\frac{a+b}{1-ab}\right) = \tan^{-1}(-7). \]

### Step 4: Calculate \( a + b \)

Calculate \( a + b \):

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

### Step 5: Calculate \( 1 - ab \)

Calculate \( 1 - ab \):

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

### Step 6: Set up the equation

Now we can set up the equation:

\[ \tan^{-1}\left(\frac{\frac{2x^2 - x + 1}{(x-1)x}}{\frac{1 - x}{x(x-1)}}\right) = \tan^{-1}(-7). \]

This simplifies to:

\[ \frac{2x^2 - x + 1}{1 - x} = -7. \]

### Step 7: Solve for \( x \)

Cross-multiply:

\[ 2x^2 - x + 1 = -7(1 - x) \implies 2x^2 - x + 1 = -7 + 7x. \]

Rearranging gives:

\[ 2x^2 - 8x + 8 = 0. \]

### Step 8: Factor the quadratic

Dividing through by 2:

\[ x^2 - 4x + 4 = 0 \implies (x - 2)^2 = 0. \]

### Step 9: Solve for \( x \)

Thus,

\[ x - 2 = 0 \implies x = 2. \]

### Final Answer

The value of \( x \) is

\[ \boxed{2}. \]

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