Question

lf aSecA, bSecB, cSecC are in $H.P$. then $a^2,b^2,c^2$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. A.G.P.

Answer 1

Answer: (A)

A.P.

$a\sec A,b\sec B,c\sec C$ are in H. P.
then $\frac{\cos A}{a},\frac{\cos B}{b},\frac{\cos C}{c}$ are in A.P.
$\frac{2\cos B}{b}=\frac{\cos A}{a}+\frac{\cos C}{c}$
$=\frac{b^2+c^2+a^2+b^2-a^2-c^2}{2abc}$
$\Rightarrow \frac{2(a^2+c^2-b^2)}{2abc}=\frac{b^2}{abc}=\frac{b}{ac}$
$a^2+c^2=2b^2$
$\therefore a^2,b^2,c^2$ are in AP.