Question

In a $\Delta$ $ABC$, the value of $\frac{(a+b+c).(b+c-b).(c+a-b).(a+b-c)}{4b^2{c}^2}$ is

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

Answer 1

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

We know that in a triangle, $2s=a+b+c$
Therefore, $\frac{(a+b+c).(b+c-b).(c+a-b).(a+b-c)}{4b^2{c}^2}$
$=\frac{\left(2s\right).(2s-2a).(2s-2b).(2s-2c)}{4b^2{c}^2}$
$=4\frac{s.(s-a)}{bc}\times \frac{(s-b).(s-c)}{bc}$
$=4{\cos }^2\frac{A}{2}\times {\sin }^2\frac{A}{2}$

$={(2\sin \frac{A}{2}\times \cos \frac{A}{2})}^2$
$={\sin }^{2}A$