Question

In a $\Delta ABC$, angles $A,B,C$ are in $A.P$. Then the value of $\underset{A\to C}{lim}\frac{\sqrt{3-4\sin A\sin C}}{|A-C|}$ is:

Answer 1

Given: $A,B,C$ are in $A.P$
Hence $B={60}^{\circ }$
So, $\cos B=\frac{1}{2}=\frac{c^2+a^2-b^2}{2ac}\Rightarrow c^2+a^2=b^2+ac\Rightarrow (a-c{)}^2=b^2-ac\Rightarrow |a-c|=\sqrt{b^2-ac}\Rightarrow |\sin A-\sin C|=\sqrt{{\sin }^{2}B-\sin A\sin C}\Rightarrow 2\cos \frac{A+C}{2}\left|\sin \frac{A-C}{2}\right|=\sqrt{\frac{3}{4}-\sin A\sin C}\Rightarrow 2\left|\sin \frac{A-C}{2}\right|=\sqrt{3-4\sin A\sin C}$
$(\because \cos \frac{A+C}{2}=\cos {60}^{\circ }=\frac{1}{2})$
Hence, $\underset{A\to C}{lim}\frac{\sqrt{3-4\sin A\sin C}}{|A-C|}=\underset{A\to C}{lim}\frac{2\sin \left|\frac{A-C}{2}\right|}{|A-C|}=1$