Question

In a $△ABC,$ if $a{\cos }^2\frac{C}{2}+c{\cos }^2\frac{A}{2}=\frac{3b}{2},$ then $a,b,c$

• A. are in A.P.
• B. are in G.P.
• C. are in H.P.
• D. satisfy $a+c=b$

Answer 1

Answer: (A)

are in A.P.

We have $\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}},$ $\cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{ac}}$ and $\cos \frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}}$
${\cos }^2\frac{A}{2}=\frac{s(s-a)}{bc}$ and ${\cos }^2\frac{C}{2}=\frac{s(s-c)}{ab}$
$\therefore a{\cos }^2\frac{C}{2}+c{\cos }^2\frac{A}{2}$
$=a\frac{s(s-c)}{ab}+c\frac{s(s-a)}{bc}$
$=\frac{s(s-c)+s(s-a)}{b}$
$=\frac{s(s-c+s-a)}{b}$
$=\frac{s(2s-a-c)}{b}$
$=\frac{s(a+b+c-a-c)}{b}$
$=\frac{sb}{b}=s=\frac{3b}{2}$(given)
$\Rightarrow 2s=3b$
$\Rightarrow a+b+c=3b$
or $a+c=2b$
$\therefore a,b,c$ are in A.P.