Question

$\text{If}\frac{Sin3B}{SinB}=(\frac{a^2-c^2}{2ac}{)}^2$ 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.

$\left(\frac{3\sin B-4{\sin }^{3}B}{\sin B}\right)={\left(\frac{a^2-C^2}{2ac}\right)}^2$
$3-4{\sin }^{2}B={\left(\frac{a^2-C^2}{2ac}\right)}^2$
$(4{\cos }^{2}B-1)={\left(\frac{a^2-C^2}{2ac}\right)}^2$
$4{\cos }^{2}B={\left(\frac{a^2+C^2}{2ac}\right)}^2$
$2\cos B=\frac{a^2+c^2}{2ac}$
$2\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+C^2}{2ac}$
$2b^2=a^2+c^2$
$\therefore a^2,b^2,c^{2}areinAP.$