Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

The expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}$ equals?

• A. ${\cos }^{2}A$
• B. ${\sin }^{2}A$
• C. $\sin 2A$
• D. $\cos 2A$

Answer 1

The correct option is B ${\sin }^{2}A$

Let $k=\frac{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}{4b^2{c}^2}$
$\Rightarrow k=\frac{2s(2s-a-a)(2s-b-b)(2s-c-c)}{4b^2{c}^2}$
Where $s$ is the semiperimeter and $2s=a+b+c$
$\Rightarrow k=\frac{16s(s-a)(s-b)(s-c)}{4b^2{c}^2}=\frac{16{△}^2}{4b^2{c}^2}$
Where $△$ is area of the triangle and $△=\frac{1}{2}bc\sin A$
$\therefore k=\frac{16{\left(\frac{1}{2}bc\sin A\right)}^2}{4b^2{c}^2}={\sin }^{2}A$