Question

In $△ABC$ angles $A,B,C$ are in $A.P.$, then $\underset{A\to C}{lim}\frac{\sqrt{3-4\sin A\sin C}}{|A-C|}$ is:

• A. $1$
• B. $2$
• C. $3$
• D. not exist

Answer 1

The correct option is A $1$

Given: In $\Delta ABC$, angles $A,B,C$ are in AP.

To find: $\underset{A\to C}{lim}\frac{\sqrt{3-4\sin A\sin C}}{|A-C|}$

$A,B,C$ are in A.P

$\Rightarrow A=B-x$ and $C=B+x$

$\Rightarrow B={60}^o\Rightarrow A+C={120}^0$

$\sin A\sin C=\frac{1}{2}\cos (A-C)-\cos (A+C)$

$=\left\{\cos \right(A-C)-\cos ({120}^o\left)\right\}\times \frac{1}{2}$

$=\frac{\cos (A-C)}{2}+\frac{1}{4}$

$\Rightarrow 3-4\sin A\sin B=3-4(\frac{\cos (A-C)}{2}+\frac{1}{4})$

$=2-2\cos (A-C)$

$=2(1-\cos (A-C\left)\right)=4{\sin }^2\left(\frac{A-C}{2}\right)$

Since, $\cos \theta =1-2{\sin }^2\frac{\theta }{2}$

$\Rightarrow \underset{A\to C}{lim}\sqrt{\frac{3-4\sin A\sin C}{|A-C|}}=\underset{A\to C}{lim}\frac{2\sin \left(\frac{A-C}{2}\right)}{|A-C|}=1$