Given,

$\Delta$ denotes the area of triangle $ABC$,

Then $\left(b\sin C+c\sin B\right)\left(b\cos C+c\cos B\right)$

$=\left(b\times \frac{c}{2R}+c\times \frac{b}{2R}\right)\left(b\cos C+c\cos B\right)$ {using sine rule}

$=\frac{2bc}{2R}\times \left(b\times \left(\frac{a^2+b^2-c^2}{2ab}\right)+c\left(\frac{a^2+c^2-b^2}{2ac}\right)\right)$

$=\frac{bc}{R}\times \left(\left(\frac{a^2+b^2-c^2}{2a}\right)+\left(\frac{a^2+c^2-b^2}{2a}\right)\right)$

$=\frac{bc}{R}\times \left(\frac{a^2+b^2-c^2+a^2+c^2-b^2}{2a}\right)$

$=\frac{bc}{R}\times \left(\frac{2a^2}{2a}\right)$

$=\frac{abc}{R}$

$=4△$ {as $△=\frac{abc}{4R}$}