Question

For a triangle $ABC$, the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}$ is equal to

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

Answer 1

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

$=\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)\times 2\times 2}{4\times bc\times bc\times 2\times 2}$
$=4\frac{s(s-a)(s-b)(s-c)}{bc\times bc}$
$=4{\cos }^{2}A/2\cdot {\sin }^{2}A/2$
$={\sin }^{2}A$