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**Hint**: We have to simplify the given equation and then using the roots given find the relation between the roots of the equation and unknown variables such as $ a,b,c,{\rm{and d}} $ .

**:**

__Complete step-by-step answer__The four non-zero real numbers satisfying the equations also known as “the roots” of the equation are given as - $ {x_1},{x_2},{x_3} $ and $ {x_4} $ .

Now considering the equation given and simplifying it we get,

$ {\tan ^{ - 1}}\dfrac{a}{x} + {\tan ^{ - 1}}\dfrac{b}{x} + {\tan ^{ - 1}}\dfrac{c}{x} + {\tan ^{ - 1}}\dfrac{d}{x} = \dfrac{\pi }{2} $

We know the formula for,

$ {\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right) $

On applying this formula in the equation we get,

$ \begin{array}{c}

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

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

\end{array} $

Now, again applying the formula for $ {\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right) $ in this equation we get,

$ {\tan ^{ - 1}}\left( {\dfrac{{\left( {\dfrac{{\left( {a + b} \right)x}}{{{x^2} - ab}}} \right) + \left( {\dfrac{{\left( {c + d} \right)x}}{{{x^2} - cd}}} \right)}}{{1 - \left( {\dfrac{{\left( {a + b} \right)x}}{{{x^2} - ab}}} \right)\left( {\dfrac{{\left( {c + d} \right)x}}{{{x^2} - cd}}} \right)}}} \right) = \dfrac{\pi }{2} $

We know that $ {\tan ^{ - 1}}\infty = \dfrac{\pi }{2} $ , so from this relation comparing the above equation we get,

$ \left( {\dfrac{{\left( {\dfrac{{\left( {a + b} \right)x}}{{{x^2} - ab}}} \right) + \left( {\dfrac{{\left( {c + d} \right)x}}{{{x^2} - cd}}} \right)}}{{\dfrac{{\left( {{x^2} - ab} \right)\left( {{x^2} - cd} \right) - \left( {a + b} \right)\left( {c + d} \right){x^2}}}{{\left( {{x^2} - ab} \right)\left( {{x^2} - cd} \right)}}}}} \right) = \infty $

This is only possible when the value of the denominator is equal to zero. So,

$ \begin{array}{c}

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

\left( {{x^2} - ab} \right)\left( {{x^2} - cd} \right) - \left( {ac + ad + bc + bd} \right){x^2} = 0\\

{x^4} - cd{x^2} - ab{x^2} + abcd - \left( {ac + ad + bc + bd} \right){x^2} = 0

\end{array} $

On rearranging this in equation form we get,

$ {x^4} - \left( {ab + bc + cd + ad + ac + bd} \right){x^2} + abcd = 0 $

Now this is an equation in $ {x^4} $ and if we solve this equation, we will get four roots.

The roots are given as $ {x_1},{x_2},{x_3} $ and $ {x_4} $ .

The sum of the roots is given as,

$ {x_1} + {x_2} + {x_3} + {x_4} = {\rm{Coefficients of}}\;{x^3} $

Since there is no term in the equation having $ {x^3} $ then, the coefficient of $ {x^3} $ is zero. So, $ {x_1} + {x_2} + {x_3} + {x_4} = 0 $

And, the product of the roots is given as-

$ \begin{array}{c}

{x_1}{x_2}{x_3}{x_4} = \dfrac{{{\rm{constant term}}}}{{{\rm{coefficient of }}{{\rm{x}}^{\rm{4}}}}}\\

{x_1}{x_2}{x_3}{x_4} = \dfrac{{abcd}}{1}\\

{x_1}{x_2}{x_3}{x_4} = abcd

\end{array} $

We can write this in a following way,

\[\sum\limits_{i = 1}^4 {{x_i}} = abcd\]

Now if we put $ x = \dfrac{1}{y} $ in the $ {x^4} $ equation we get,

$ \dfrac{1}{{{y^4}}} - \left( {ab + bc + cd + ad + ac + bd} \right)\dfrac{1}{{{y^2}}} + abcd = 0 $

Now on solving this we get,

$ abcd{y^4} - \left( {ab + bc + cd + ad + ac + bd} \right){y^2} + 1 = 0 $

This also reduces to an equation in $ {y^4} $ and its roots are also given as- $ {y_1},{y_2},{y_3}{\rm{ and }}{y_4} $ .

Now we know that the sum of the roots is,

$ {y_1} + {y_2} + {y_3}{\rm{ + }}{y_4} = {\rm{ Coefficient of }}{y^3} $

So, the coefficient of $ {y^3} $ is zero, that is,

$ {y_1} + {y_2} + {y_3}{\rm{ + }}{y_4} = {\rm{ 0}} $

And since $ x = \dfrac{1}{y} $ , we can write,

$ {y_1} = \dfrac{1}{{{x_1}}} $ , $ {y_2} = \dfrac{1}{{{x_2}}} $ , $ {y_3} = \dfrac{1}{{{x_3}}} $ and $ {y_4} = \dfrac{1}{{{x_4}}} $ substituting these values, we get,

\[\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \dfrac{1}{{{x_3}}} + \dfrac{1}{{{x_4}}} = 0\]

We can write this in a following way,

$ \sum\limits_{i = 1}^4 {\dfrac{1}{{{x_i}}}} = 0 $

**So, the correct answer is “Option C”.**

**Note**: The formula to calculate the sum of the roots and the product of the roots is obtained in the same way for any equation no matter what the order of the equation is. For example, for a quadratic equation given by –

$ a{x^2} + bx + c = 0 $

There are two roots of this equation, these roots are $ {x_1} $ and $ {x_2} $ .

Now, the sum of the roots will be,

$ {x_1} + {x_2} = - \dfrac{b}{a} $

Where $ b $ is the coefficient of $ x $ , and $ a $ is the coefficient of $ {x^2} $ .

And, the product of the roots is,

$ {x_1}{x_2} = \dfrac{c}{a} $

Where, $ c $ is a constant term.

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