Question

In a triangle $ABC,$ if $\cos (A-C)\cos B+\cos 2B=0$ then $a^2,b^2,c^2$ are in

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

Answer 1

The correct option is C A.P

$\cos (A-C)\cos B=-\cos 2B$
$\Rightarrow \cos (A-C)\cos (\pi -(A+C))=-\cos 2B$
$\Rightarrow -\cos (A-C)\cos (A+C)=-\cos 2B$
$\Rightarrow \cos (A-C)\cos (A+C)=\cos 2B$
$\Rightarrow {\cos }^{2}A-{\sin }^{2}C=1-2{\sin }^{2}B$
$\Rightarrow 1-{\sin }^{2}A-{\sin }^{2}C=1-2{\sin }^{2}B$
$\Rightarrow {\sin }^{2}A+{\sin }^{2}C=2{\sin }^{2}B$
$\Rightarrow a^2+c^2=2b^2$ by sine rule.
Hence $a^2,b^2,c^2$ are in A.P.