Question

The expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}$ is equal to

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

Answer 1

The correct option is B ${\sin }^{2}A$

We have,

$\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}$

We know that, In a triangle

$2S=a+b+c$

$\frac{2s(a+b+c-2a)(c+a+b-2b)(a+b+c-2c)}{4b^2{c}^2}$

$=\frac{2s(2s-2a)(2s-2b)(2s-2c)}{4b^2{c}^2}$

$=\frac{16s(s-a)(s-b)(s-c)}{4b^2{c}^2}$

$=\frac{16{\Delta }^2}{4b^2{c}^2}$

$=\frac{4{\Delta }^2}{b^2{c}^2}$

We know that, $\Delta =\frac{1}{2}bc\sin A$

$\frac{4}{b^2{c}^2}{\left(\frac{1}{2}bc\sin A\right)}^2$

$=\frac{4}{b^2{c}^2}\times \frac{1}{4}{b}^2{c}^2{\sin }^{2}A$

$={\sin }^{2}A$

Hence, this is the answer.