Question

In a $△ABC,(b-c{)}^2+4bc{\sin }^2\frac{A}{2}$ is equal to

• A. $a^2$
• B. $2a$
• C. $(\frac{b+c}{a}{)}^2$
• D. $noneofthese$

Answer 1

The correct option is A $a^2$

$(b-c{)}^2+4bc{\sin }^2\frac{A}{2}$

$\Rightarrow b^2+c^2-2bc+4bc{\sin }^2\frac{A}{2}+4bc{\cos }^2\frac{A}{2}-4bc{\cos }^2\frac{A}{2}$

$\Rightarrow b^2+c^2-2bc-\frac{4bc{\cos }^{2}A/2}{2}$

$\Rightarrow b^2+c^2-2bc(1-2{\cos }^{2}A/2)\Rightarrow B^2+c^2+2bc\cos A$

$A^2=b^2+c^2+2bc\cos A$