Question

In any $∆ABC,\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}$ equals:

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

Answer 1

The correct option is D

${\sin }^{2}A$

Explanation for the correct option.

Finding the value of the expression:

As per the Heron's Formula, in $∆ABC$, $a+b+c=2s$. So,

$\begin{array}{rcl}\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}& =& \frac{\left(2s\right)(2s-2a)(2s-2b)(2s-2c)}{4b^2{c}^2}\\ & =& \frac{16s\left(s-a\right)\left(s-b\right)\left(s-c\right)}{4b^2{c}^2}\\ & =& \frac{4\left[s\left(s-a\right)\left(s-b\right)\left(s-c\right)\right]}{bc\times bc}\\ & =& \frac{4{\left[∆\right]}^2}{bc\times bc}\left[\mathrm{By}\mathrm{Area}\mathrm{of}∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\right]\\ & =& \frac{4{\left[{\displaystyle \frac{1}{2}}bc\sin A\right]}^2}{bc\times bc}\left[\mathrm{By}\mathrm{Area}\mathrm{of}∆=\frac{1}{2}\mathrm{bc}\sin A\right]\\ & =& {\sin }^{2}A\end{array}$

Hence, option D is correct.