Question

In $△ABC$, if $\cos A+\cos C=4{\sin }^2\frac{B}{2}$, then $a,b,c$ (sides of the triangle) are in

• A. A.P
• B. G.P
• C. H.P
• D. None of these

Answer 1

The correct option is A A.P

$\cos A+\cos C=4{\sin }^2\frac{B}{2}$
$\Rightarrow 2\cos \left(\frac{A+C}{2}\right)+\cos \left(\frac{A-C}{2}\right)=4{\sin }^2\frac{B}{2}\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\sin \frac{B}{2}\Rightarrow \cos \frac{A}{2}\cos \frac{C}{2}+\sin \frac{A}{2}\sin \frac{C}{2}=2\sin \frac{B}{2}\Rightarrow \sqrt{\frac{s(s-a)}{bc}}\sqrt{\frac{s(s-c)}{ab}}+\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-b)}{ab}}=2\sqrt{\frac{(s-a)(s-c)}{ac}}$
$\Rightarrow \frac{s}{b}+\frac{s-b}{b}=2\Rightarrow a+c=2b$
So, they are in A.P.