The expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is :
Answer & explanation
Correct answer: option 4
Let, $E=\frac{2s.2(s-a)2.(s-b)2.(s-c)}{4b^2c^2}=\frac{4}{b^2c^2}Δ^2=\frac{4}{b^2c^2}(\frac{1}{2}bc\sin A)^2=\sin^2 A$
The expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is :
Correct answer: option 4
Let, $E=\frac{2s.2(s-a)2.(s-b)2.(s-c)}{4b^2c^2}=\frac{4}{b^2c^2}Δ^2=\frac{4}{b^2c^2}(\frac{1}{2}bc\sin A)^2=\sin^2 A$