If `tan^(-1)(x+2)+tan^(-1)(x-2)-tan^(-1)((1)/(2))=0`, them one of the values of x is equal to

Options

A. `-1`
B. 5
C. `(1)/(2)`
D. 1

Explanation

To solve the equation \( \tan^{-1}(x+2) + \tan^{-1}(x-2) - \tan^{-1}\left(\frac{1}{2}\right) = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
We can rearrange the equation to isolate the inverse tangent terms:
\[ \tan^{-1}(x+2) + \tan^{-1}(x-2) = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 2: Apply the formula for the sum of inverse tangents
Using the identity \( \tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A + B}{1 - AB}\right) \), we can let \( A = x + 2 \) and \( B = x - 2 \):
\[ \tan^{-1}\left(\frac{(x+2) + (x-2)}{1 - (x+2)(x-2)}\right) = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 3: Simplify the left-hand side
Calculating the numerator and denominator:
- Numerator: \( (x + 2) + (x - 2) = 2x \)
- Denominator: \( 1 - (x^2 - 4) = 1 - x^2 + 4 = 5 - x^2 \)

Thus, we have:
\[ \tan^{-1}\left(\frac{2x}{5 - x^2}\right) = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 4: Set the arguments equal
Since the inverse tangent functions are equal, we can set their arguments equal to each other:
\[ \frac{2x}{5 - x^2} = \frac{1}{2} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ 2(2x) = 1(5 - x^2) \]
This simplifies to:
\[ 4x = 5 - x^2 \]

### Step 6: Rearrange into standard form
Rearranging this equation gives:
\[ x^2 + 4x - 5 = 0 \]

### Step 7: Factor the quadratic equation
We can factor this quadratic:
\[ (x + 5)(x - 1) = 0 \]

### Step 8: Solve for x
Setting each factor to zero gives us the possible solutions:
\[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Conclusion
Thus, one of the values of \( x \) is \( -5 \) and the other value is \( 1 \).