Q103 / 160 · TS EAMCET 2014 · Mathematics
In any
$$
\triangle A B C, \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2}
$$
equals to
- A$\sin ^2 B$
- B$\cos ^2 A$
- C$\cos ^2 B$
- Correct answerD$\sin ^2 A$
Solution
$\begin{aligned} & \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} \\ & =\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^2 c^2} \\ & =\frac{16[s(s-a)(s-b)(s-c)]}{4 b^2 c^2} \quad(\because a+b+c=2 s) \\ & =\frac{4 \Delta^2}{b^2 c^2}=\left(\frac{2 \Delta}{b c}\right)^2=\sin ^2 A\left[\because \sin A=\frac{2 \Delta}{b c}\right]\end{aligned}$
Hence, the correct answer is option: 'D'.