Q54 / 120 · VITEEE 2009 · Mathematics
In a $\triangle \mathrm{ABC}$
$\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^{2} c^{2}}$
equals
- A$\cos ^{2} \mathrm{~A}$
- B$\cos ^{2} \mathrm{~B}$
- Correct answerC$\sin ^{2} \mathrm{~A}$
- D$\sin ^{2} \mathrm{~B}$
Solution
We know that, $2 s=a+b+c$
$$
\begin{array}{l}
\therefore \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^{2} c^{2}} \\
=\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^{2} c^{2}} \\
=4 \frac{s(s-a)}{b c} \times \frac{(s-b)(s-c)}{b c} \\
=4 \cos ^{2} \frac{A}{2} \times \sin ^{2} \frac{A}{2}=\sin ^{2} A
\end{array}
$$
Hence, the correct answer is option: 'C'.