Q69 / 160 · AP EAMCET 2009 · Mathematics

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

  • A
    $\cos ^2 A$
  • B
    $\cos ^2 B$
  • Correct answer
    C
    $\sin ^2 A$
  • D
    $\sin ^2 B$

Solution

We know that, $2 s=a+b+c$ $\begin{aligned} & \therefore \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} \\ & =\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^2 c^2} \\ & =4 \frac{s(s-a)}{b c} \times \frac{(s-b)(s-c)}{b c} \\ & =4 \cos ^2 \frac{A}{2} \times \sin ^2 \frac{A}{2}=\sin ^2 A \\ & \end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions