Question

In a \(\triangle ABC\)
\[ \frac{(a+b-c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2} \] equals

• A. \(\cos^2A\)
• B. \(\cos^2B\)
• C. \(\sin^2A\)
• D. \(\sin^2B\)

Hint:

Remember: \(16\Delta^2=(a+b+c)(a+b-c)(b+c-a)(c+a-b)\). Substitute into \(\sin^2A=\frac{4\Delta^2}{b^2c^2}\).

Answer 1

The Correct Option is C

Step 1: Recognize standard triangle identity.
Expression resembles form:
\[ \frac{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}{16b^2c^2} \]
which is known to equal \(\sin^2A\).
Step 2: Use formula for \(\sin A\) in terms of sides.
\[ \sin A = \frac{2\Delta}{bc} \Rightarrow \sin^2A=\frac{4\Delta^2}{b^2c^2} \]
Step 3: Use Heron’s formula.
\[ 16\Delta^2=(a+b+c)(a+b-c)(b+c-a)(c+a-b) \]
So:
\[ \sin^2A=\frac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{4b^2c^2} \]
Step 4: Match with given expression.
Hence expression equals \(\sin^2A\).
Final Answer:
\[ \boxed{\sin^2A} \]