In a $$\Delta \,ABC,\frac{{\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)}}{{4{b^2}{c^2}}}$$           equals

Solution :
We know that, $$2s = a + b + c$$
$$\eqalign{ & \therefore \frac{{\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)}}{{4{b^2}{c^2}}} \cr & = \frac{{2s\left( {2s - 2a} \right)\left( {2s - 2b} \right)\left( {2s - 2c} \right)}}{{4{b^2}{c^2}}} \cr & = 4\frac{{s\left( {s - a} \right)}}{{bc}} \times \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}} \cr & = 4{\cos ^2}\frac{A}{2} \times {\sin ^2}\frac{A}{2} \cr & = {\sin ^2}A \cr} $$