Question

In any $△ABC$, $\frac{(a+b+c)(b+c-a)(a+b-c)(c+b-a)}{4b^2{c}^2}=$

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

Answer 1

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

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

$s=\frac{a+b+c}{2}$

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

$\Rightarrow 4\times {\left(\sqrt{\frac{(s-b)(s-c)}{bc}}\right)}^2{\left(\sqrt{\frac{s(s-a)}{bc}}\right)}^2$

$=4{\sin }^2\frac{A}{2}{\cos }^2\frac{A}{2}$

$={\sin }^{2}A$