In $\vartriangle PQR,\,\angle R = \dfrac{\pi }{2}$.If $\tan \left( {\dfrac{P}{2}} \right)$ and $\tan \left( {\dfrac{Q}{2}} \right)$ are the roots of $a{x^2} + bx + c = 0,a \ne 0$ then
A. $a + b = c$
B. $b + c = a$
C. $a + c = b$
D. $b = c$

Hint: Given, In $\vartriangle PQR,\,\angle R = \dfrac{\pi }{2}$. And $\tan \left( {\dfrac{P}{2}} \right)$ and $\tan \left( {\dfrac{Q}{2}} \right)$ are the roots of $a{x^2} + bx + c = 0,a \ne 0$. First, we will find the measurement of the other two angles. Then we find the sum and product of the roots of the given equation. Then we will use $\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\,\tan b\;}}$ to find the required relation.

Formula Used:
$\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\,\tan b\;}}$

Complete step by step Solution:
 Given, In $\vartriangle PQR,\,\angle R = \dfrac{\pi }{2}$
A triangle's total angles equal the straight angle (180 degrees, radians, two right angles, or a half-turn) in a Euclidean space. A triangle has three angles, one at each vertex, delimited by a pair of neighbouring sides.
$\angle P + \angle Q + \angle R = \pi $
Given, $\angle R = \dfrac{\pi }{2}$
$P + Q = \pi - \dfrac{\pi }{2}$
$P + Q = \dfrac{\pi }{2}$
Dividing both sides by two.
$\dfrac{{P + Q}}{2} = \dfrac{\pi }{4}$
$\dfrac{P}{2} + \dfrac{Q}{2} = \dfrac{\pi }{4}$
$\tan \left( {\dfrac{P}{2}} \right)$ and $\tan \left( {\dfrac{Q}{2}} \right)$ are the roots of $a{x^2} + bx + c = 0$
The solutions of the quadratic equation $a{x^2} + bx + c = 0$ serve as the quadratic equation's roots. They are the values of the variable $x$ that the equation requires. The x-coordinates of the x-intercepts of a quadratic function are the roots of the function. A quadratic equation can only have a maximum of two roots because its degree is 2. We can use a variety of techniques to locate the roots of quadratic equations as follow:
Quadratic formula
Graphing
Factoring
Completing the square
We know sum of roots of quadratic equation is $\dfrac{{ - b}}{a}$
And product of quadratics equation is $\dfrac{c}{a}$
$\tan \left( {\dfrac{P}{2}} \right) + \tan \left( {\dfrac{Q}{2}} \right) = \dfrac{{ - b}}{a}$
$\tan \left( {\dfrac{P}{2}} \right).\tan \left( {\dfrac{Q}{2}} \right) = \dfrac{c}{a}$
We know that $\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\,\tan b\;}}$
Let $a = \dfrac{P}{2},\,b = \dfrac{Q}{2}$
$\tan \left( {\dfrac{P}{2} + \dfrac{Q}{2}} \right) = \tan \left( {\dfrac{\pi }{4}} \right)$
$\tan \left( {\dfrac{P}{2} + \dfrac{Q}{2}} \right) = \dfrac{{\tan \dfrac{P}{2} + \tan \dfrac{Q}{2}}}{{1 - \tan \dfrac{P}{2}\,\tan \dfrac{Q}{2}\;}}$
After applying the sum and products of roots of the quadratic equation
$\tan \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\dfrac{{ - b}}{a}}}{{1 - \dfrac{c}{a}}}$
$1 = \dfrac{{\dfrac{{ - b}}{a}}}{{\dfrac{{a - c}}{a}}}$
After simplification
\[1 = \dfrac{{ - b}}{{a - c}}\]
$a - c = - b$
$a + b = c$

Hence, the correct option is A.

Note: Students should pay attention while finding the remaining angle then use the correct formula while finding the sum and product of quadratic equations. Use formula $\tan (a + b)$ correctly then put correct values in order to get the correct required relation.