Question

Prove that $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}={\sin }^{2}A$

Answer 1

$\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}$

substituting a,b,c in form of "s" and simplifying

$=\frac{2s.2(s-a).2(s-b).2(s-c)}{4b^2{c}^2}$

$=\frac{4{\Delta }^2}{b^2{c}^2}$ $=\frac{4}{b^2{c}^2}{\left(\frac{1}{2}bc\sin A\right)}^2={\sin }^{2}A$