We have to find the value of $\frac{{\cos }^2\left(\frac{B-C}{2}\right)}{(b+c{)}^2}+\frac{{\sin }^2\left(\frac{B-C}{2}\right)}{(b-c{)}^2}$,

We know that, in a triangle, $a=2R\sin A, b=2R\sin B \mathrm{and} c=2R\sin C$

Substituting them,

$\frac{{\cos }^2\left(\frac{B-C}{2}\right)}{(b+c{)}^2}+\frac{{\sin }^2\left(\frac{B-C}{2}\right)}{(b-c{)}^2}=\frac{{\cos }^2\left(\frac{B-C}{2}\right)}{{\left(2R· \sin B+2R· \sin C\right)}^2}+\frac{{\sin }^2\left(\frac{B-C}{2}\right)}{2R· \sin B-2R· \sin C{\mathbf{)}}^{\mathbf{2}}}$

$=\frac{1}{4r^2}\left[\frac{{\cos }^2\left(\frac{B-C}{2}\right)}{{\left( \sin B+ \sin C\right)}^2}+\frac{{\sin }^2\left(\frac{B-C}{2}\right)}{{\left(\sin B-\sin C\right)}^{\mathbf{2}}}\right]=\frac{1}{4r^2}\left[\frac{{\cos }^2\left(\frac{B-C}{2}\right)}{{\left(2\sin {\displaystyle \frac{B+C}{2}}·\cos {\displaystyle \frac{B-C}{2}}\right)}^2}+\frac{{\sin }^2\left(\frac{B-C}{2}\right)}{{\left(-2\cos {\displaystyle \frac{B+C}{2}}·\sin {\displaystyle \frac{B-C}{2}}\right)}^{\mathbf{2}}}\right]=$$=\frac{1}{16r^2}\left[\frac{1}{{\sin }^2{\displaystyle \frac{B+C}{2}}}+\frac{1}{{\cos }^2{\displaystyle \frac{B+C}{2}}}\right]=$$=\frac{1}{4r^2}\left(\frac{{\cos }^2{\displaystyle \frac{B+C}{2}}+{\sin }^2{\displaystyle \frac{B+C}{2}}}{4{\sin }^2{\displaystyle \frac{B+C}{2}·{\cos }^2\frac{{\displaystyle B+C}}{{\displaystyle 2}}}}\right)=\frac{1}{4r^2}\left(\frac{1}{4{\sin }^2{\displaystyle \frac{B+C}{2}·{\cos }^2\frac{{\displaystyle B+C}}{{\displaystyle 2}}}}\right)=\frac{1}{4r^2}\left(\frac{1}{{\sin }^2\left(B+C\right)}\right)$

As, $A+B+C=\mathrm{\pi }$$=\frac{1}{{\left(2R\sin \left(B+C\right)\right)}^2}=\frac{1}{{\left(2R\sin \left(\mathrm{\pi }-\mathrm{A}\right)\right)}^2}=\frac{1}{{\left(2R\mathrm{sinA}\right)}^2}=\frac{1}{a^2}$