Question

In a \( \Delta ABC \), \( \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^{2}c^{2}} \) equals

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

Hint:

Heron's Formula logic: $16s(s-a)(s-b)(s-c) = 16\Delta^2$.

Answer 1

The Correct Option is C

Step 1: Semi-perimeter
Let $2s = a+b+c$. Then $b+c-a = 2(s-a)$, $c+a-b = 2(s-b)$, and $a+b-c = 2(s-c)$.
Step 2: Substitution
$\frac{2s \cdot 2(s-a) \cdot 2(s-b) \cdot 2(s-c)}{4b^2c^2} = 4 \frac{s(s-a)}{bc} \cdot \frac{(s-b)(s-c)}{bc}$.
Step 3: Half-angle formulas
$= 4 \cos^2(A/2) \sin^2(A/2)$.
Step 4: Simplify
$= (2 \sin(A/2) \cos(A/2))^2 = \sin^2 A$.
Final Answer: (C)