Question

In a $\Delta A B C \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^{2} c^{2}}$ equals

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

Answer 1

Answer: (C)

$\sin ^{2} A$

Answer 2

The correct answer is C:$sin^2A$
We know that, $2 s=a+b+c$
$\therefore \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^{2} c^{2}}$
$=\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^{2} c^{2}}$
$=4 \frac{s(s-a)}{b c} \times \frac{(s-a)(s-c)}{b c}$
$=4 \cos ^{2} \frac{A}{2} \times \sin ^{2} \frac{A}{2}$
$=\sin ^{2} A$