The background to this question is here.
In the expression:
$2 \tan^{-1} \left[ \dfrac{a-20}{c-20}. \dfrac{\sqrt{(a-20)^2+(b-20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(a-20)^2+(b-20)^2}-(b-20)} \right]$ within $\dfrac{\partial E_{x}}{\partial a}$,
if we consider the 8 variations of $a \pm 20$,$b \pm 20$,$c \pm 20$, we get 8 terms.
In the expression:
$2 \tan^{-1} \left[ \dfrac{b-20}{a-20}. \dfrac{\sqrt{(b-20)^2+(c-20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(b-20)^2+(c-20)^2}-(c-20)} \right]$ within $\dfrac{\partial E_{y}}{\partial b}$,
if we consider the 8 variations of $a \pm 20$,$b \pm 20$,$c \pm 20$, we get 8 terms.
In the expression:
$2 \tan^{-1} \left[ \dfrac{c-20}{b-20}. \dfrac{\sqrt{(c-20)^2+(a-20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(c-20)^2+(a-20)^2}-(a-20)} \right]$ within $\dfrac{\partial E_{z}}{\partial c}$,
if we consider the 8 variations of $a \pm 20$,$b \pm 20$,$c \pm 20$, we get 8 terms.
Since $\vec{\nabla}.\vec{E}=\dfrac{\partial E_{x}}{\partial a}+\dfrac{\partial E_{y}}{\partial b}+\dfrac{\partial E_{z}}{\partial c}=0$, I need to get the sum of the following $8 + 8 + 8 = 24$ terms as zero :
\begin{align} \vec{\nabla}.\vec{E} &=\dfrac{\partial E_{x}}{\partial a}+\dfrac{\partial E_{y}}{\partial b}+\dfrac{\partial E_{z}}{\partial c}\\ &=2 \tan^{-1} \left[ \dfrac{a-20}{c-20}. \dfrac{\sqrt{(a-20)^2+(b-20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(a-20)^2+(b-20)^2}-(b-20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{a-20}{c+20}. \dfrac{\sqrt{(a-20)^2+(b-20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c+20)^2}}{\sqrt{(a-20)^2+(b-20)^2}-(b-20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{a-20}{c-20}. \dfrac{\sqrt{(a-20)^2+(b+20)^2}-\sqrt{(a-20)^2+(b+20)^2+(c-20)^2}}{\sqrt{(a-20)^2+(b+20)^2}-(b+20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{a-20}{c+20}. \dfrac{\sqrt{(a-20)^2+(b+20)^2}-\sqrt{(a-20)^2+(b+20)^2+(c+20)^2}}{\sqrt{(a-20)^2+(b+20)^2}-(b+20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{a+20}{c-20}. \dfrac{\sqrt{(a+20)^2+(b-20)^2}-\sqrt{(a+20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(a+20)^2+(b-20)^2}-(b-20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{a+20}{c+20}. \dfrac{\sqrt{(a+20)^2+(b-20)^2}-\sqrt{(a+20)^2+(b-20)^2+(c+20)^2}}{\sqrt{(a+20)^2+(b-20)^2}-(b-20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{a+20}{c-20}. \dfrac{\sqrt{(a+20)^2+(b+20)^2}-\sqrt{(a+20)^2+(b+20)^2+(c-20)^2}}{\sqrt{(a+20)^2+(b+20)^2}-(b+20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{a+20}{c+20}. \dfrac{\sqrt{(a+20)^2+(b+20)^2}-\sqrt{(a+20)^2+(b+20)^2+(c+20)^2}}{\sqrt{(a+20)^2+(b+20)^2}-(b+20)} \right]\\ +\\ &+2 \tan^{-1} \left[ \dfrac{b-20}{a-20}. \dfrac{\sqrt{(b-20)^2+(c-20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(b-20)^2+(c-20)^2}-(c-20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{b-20}{a+20}. \dfrac{\sqrt{(b-20)^2+(c-20)^2}-\sqrt{(a+20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(b-20)^2+(c-20)^2}-(c-20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{b-20}{a-20}. \dfrac{\sqrt{(b-20)^2+(c+20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c+20)^2}}{\sqrt{(b-20)^2+(c+20)^2}-(c+20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{b-20}{a+20}. \dfrac{\sqrt{(b-20)^2+(c+20)^2}-\sqrt{(a+20)^2+(b-20)^2+(c+20)^2}}{\sqrt{(b-20)^2+(c+20)^2}-(c+20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{b+20}{a-20}. \dfrac{\sqrt{(b+20)^2+(c-20)^2}-\sqrt{(a-20)^2+(b+20)^2+(c-20)^2}}{\sqrt{(b+20)^2+(c-20)^2}-(c-20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{b+20}{a+20}. \dfrac{\sqrt{(b+20)^2+(c-20)^2}-\sqrt{(a+20)^2+(b+20)^2+(c-20)^2}}{\sqrt{(b+20)^2+(c-20)^2}-(c-20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{b+20}{a-20}. \dfrac{\sqrt{(b+20)^2+(c+20)^2}-\sqrt{(a-20)^2+(b+20)^2+(c+20)^2}}{\sqrt{(b+20)^2+(c+20)^2}-(c+20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{b+20}{a+20}. \dfrac{\sqrt{(b+20)^2+(c+20)^2}-\sqrt{(a+20)^2+(b+20)^2+(c+20)^2}}{\sqrt{(b+20)^2+(c+20)^2}-(c+20)} \right]\\ +\\ &+2 \tan^{-1} \left[ \dfrac{c-20}{b-20}. \dfrac{\sqrt{(c-20)^2+(a-20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(c-20)^2+(a-20)^2}-(a-20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{c-20}{b+20}. \dfrac{\sqrt{(c-20)^2+(a-20)^2}-\sqrt{(a-20)^2+(b+20)^2+(c-20)^2}}{\sqrt{(c-20)^2+(a-20)^2}-(a-20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{c-20}{b-20}. \dfrac{\sqrt{(c-20)^2+(a+20)^2}-\sqrt{(a+20)^2+(b-20)^2+(c-20)^2}}{\sqrt{(c-20)^2+(a+20)^2}-(a+20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{c-20}{b+20}. \dfrac{\sqrt{(c-20)^2+(a+20)^2}-\sqrt{(a+20)^2+(b+20)^2+(c-20)^2}}{\sqrt{(c-20)^2+(a+20)^2}-(a+20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{c+20}{b-20}. \dfrac{\sqrt{(c+20)^2+(a-20)^2}-\sqrt{(a-20)^2+(b-20)^2+(c+20)^2}}{\sqrt{(c+20)^2+(a-20)^2}-(a-20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{c+20}{b+20}. \dfrac{\sqrt{(c+20)^2+(a-20)^2}-\sqrt{(a-20)^2+(b+20)^2+(c+20)^2}}{\sqrt{(c+20)^2+(a-20)^2}-(a-20)} \right]\\ &+2 \tan^{-1} \left[ \dfrac{c+20}{b-20}. \dfrac{\sqrt{(c+20)^2+(a+20)^2}-\sqrt{(a+20)^2+(b-20)^2+(c+20)^2}}{\sqrt{(c+20)^2+(a+20)^2}-(a+20)} \right]\\ &-2 \tan^{-1} \left[ \dfrac{c+20}{b+20}. \dfrac{\sqrt{(c+20)^2+(a+20)^2}-\sqrt{(a+20)^2+(b+20)^2+(c+20)^2}}{\sqrt{(c+20)^2+(a+20)^2}-(a+20)} \right]\\ \end{align}
How is it done? Shall I use any formula of inverse trigonometric functions?
