Question

In a triangle $ABC$, the angles are in A.P., then
$\underset{A\to C}{Lt}\frac{\sqrt{3-4\sin A\sin C}}{|A-C|}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$1$

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

$b^2=a^2+c^2-ac$ $a-c=\pm \sqrt{b^2-ac}$

$b^2-ac=(a-c{)}^2$ $2\sin \frac{A-C}{2}.\cos \frac{(A+C)}{2}=\pm \sqrt{\frac{3}{4}-\sin A\sin C}$

$2\sin \left(\frac{A-C}{2}\right)=\sqrt{3-4\sin A\sin C}$

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