Question

$\Delta ABC$ with sides $a,b,c$ opposite to the angles $A,B,C$ satisfies the equation
${\log }_2\left(2\cos \frac{C}{2}\cos \frac{A-B}{2}\right)-{\log }_2\left(\sin \right(A+B\left)\right)=1$, then the value of $(a^2+b^2-4c^2+2ab)$ is

Answer 1

${\log }_2\left(2\cos \frac{C}{2}\cos \frac{A-B}{2}\right)-{\log }_2\left(\sin \right(A+B\left)\right)=1$
$\Rightarrow {\log }_{2}2+{\log }_2\left(\cos \frac{C}{2}\cos \frac{A-B}{2}\right)-{\log }_2\left(\sin \right(A+B\left)\right)=1$
$\Rightarrow {\log }_2\left(\cos \frac{\pi -(A+B)}{2}\cos \frac{A-B}{2}\right)={\log }_2\left(\sin \right(A+B\left)\right)$
$\Rightarrow \sin \frac{A+B}{2}\cos \frac{A-B}{2}=\sin (A+B)$
$\Rightarrow \frac{\sin A+\sin B}{2}=\sin (\pi -C)$
$\Rightarrow \sin A+\sin B=2\sin C$
​​​​​​​$\Rightarrow \frac{a}{2R}+\frac{b}{2R}=\frac{2c}{2R}$
​​​​​​​$\Rightarrow a+b=2c$
$a^2+b^2-4c^2+2ab=(a+b{)}^2-(2c{)}^2=0$
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