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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
Options:
$$
\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2}
$$
equals
Solution:
1250 Upvotes
Verified Answer
The correct answer is:
$\sin ^2 A$
We know that, $2 s=a+b+c$
$\begin{aligned}
& \therefore \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} \\
& =\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^2 c^2} \\
& =4 \frac{s(s-a)}{b c} \times \frac{(s-b)(s-c)}{b c} \\
& =4 \cos ^2 \frac{A}{2} \times \sin ^2 \frac{A}{2}=\sin ^2 A \\
&
\end{aligned}$
$\begin{aligned}
& \therefore \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} \\
& =\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^2 c^2} \\
& =4 \frac{s(s-a)}{b c} \times \frac{(s-b)(s-c)}{b c} \\
& =4 \cos ^2 \frac{A}{2} \times \sin ^2 \frac{A}{2}=\sin ^2 A \\
&
\end{aligned}$
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