Question

In any $\Delta ABC,$ $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}$ is equal to

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

Answer 1

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

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2}bc\sin A$
Taking square of both sides,
$\Rightarrow s(s-a)(s-b)(s-c)=\frac{1}{4}{b}^2{c}^2{\sin }^{2}A$
Mutiplying and Dividing each term by $2$
$\Rightarrow \frac{2s(2s-2a)(2s-2b)(2s-2c)}{16}=\frac{1}{4}{b}^2{c}^2{\sin }^{2}A$
$\Rightarrow \frac{2s(2s-2a)(2s-2b)(2s-2c)}{4b^2{c}^2}={\sin }^{2}A$