Let `x_1,x_2,x_3,x_4` be four non zero numbers satisfying the equation `tan^-1 (a/x)+tan^-1(b/x)+tan^-1(c/x)+tan^-1(d/x)=pi/2` then which ofthe following relation(s) hold good?

Options

A. `x_(1)+x_(2)+x_(3)+x_(4)=a+b+c+d`
B. `(1)/(x_(1))+(1)/(x_(2))+(1)/(x_(3))+(1)/(x_(4))=0`
C. `x_(1)x_(2)x_(3)x_(4)=abcd`
D. `(x_(2)+x_(3)+x_(4))(x_(3)+x_(4)+x_(1))(x_(1)+x_(2)+x_(3))=abcd`

Explanation

To solve the problem, we need to analyze the given equation:

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

### Step 1: Use the property of inverse tangent

We know that:

\[ \tan^{-1}(u) + \tan^{-1}(v) = \tan^{-1}\left(\frac{u + v}{1 - uv}\right) \quad \text{if } uv < 1 \]

We can apply this property to the first two terms:

Let \( u = \frac{a}{x} \) and \( v = \frac{b}{x} \):

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

### Step 2: Combine the next two terms

Now, we apply the same property to the last two terms:

Let \( u = \frac{c}{x} \) and \( v = \frac{d}{x} \):

\[ \tan^{-1}\left(\frac{c}{x}\right) + \tan^{-1}\left(\frac{d}{x}\right) = \tan^{-1}\left(\frac{\frac{c}{x} + \frac{d}{x}}{1 - \frac{c}{x} \cdot \frac{d}{x}}\right) = \tan^{-1}\left(\frac{c + d}{x} \cdot \frac{x^2}{x^2 - cd}\right) \]

### Step 3: Combine all terms

Now we combine both results:

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

Using the property of inverse tangent again, we have:

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

### Step 4: Set the denominator to zero

For the above equation to hold true, the denominator must be zero:

\[ 1 - \left(\frac{(a + b)(c + d)}{(x^2 - ab)(x^2 - cd)}\right) = 0 \]

This leads us to:

\[ (a + b)(c + d) = (x^2 - ab)(x^2 - cd) \]

### Step 5: Expand and rearrange

Expanding the right-hand side gives:

\[ x^4 - (ab + cd)x^2 + abcd = (a + b)(c + d) \]

This leads to a polynomial equation in \( x \):

\[ x^4 - (ab + cd)x^2 + (abcd - (a + b)(c + d)) = 0 \]

### Step 6: Analyze the roots

This is a quartic equation, which can have four roots \( x_1, x_2, x_3, x_4 \).

Using Vieta's formulas, we can find:

1. The sum of the roots \( x_1 + x_2 + x_3 + x_4 = 0 \) (since the coefficient of \( x^3 \) is zero).
2. The product of the roots \( x_1 x_2 x_3 x_4 = abcd \).

### Conclusion

Thus, the relations that hold good are:

1. \( x_1 + x_2 + x_3 + x_4 = 0 \)
2. \( x_1 x_2 x_3 x_4 = abcd \)