In a $ riangle A B C$ $$ \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} $$ equals

Question

In a $	riangle A B C$ $$ \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} $$ equals

Aicharya · Accepted Answer

We know that, $2 s=a+b+c$ $\begin{aligned} & 	herefore \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} \ & =\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^2 c^2} \ & =4 \frac{s(s-a)}{b c} 	imes \frac{(s-b)(s-c)}{b c} \ & =4 \cos ^2 \frac{A}{2} 	imes \sin ^2 \frac{A}{2}=\sin ^2 A \ & \end{aligned}$

In a $\triangle A B C$ $$ \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} $$ equals

Solution