If in A B C, 2 b^2=a^2+c^2, then 3 B B = - Vidyadip

MCQ

If in $\triangle A B C, 2 b^2=a^2+c^2$, then $\frac{\sin 3 B}{\sin B}=$

A
$\frac{ c ^2- a ^2}{2 ca }$
B
$\frac{c^2-a^2}{c a}$
C
$\left(\frac{ c ^2- a ^2}{ ca }\right)^2$
✓
$\left(\frac{ c ^2- a ^2}{2 ca }\right)^2$

✓

Answer

Correct option: D.

$\left(\frac{ c ^2- a ^2}{2 ca }\right)^2$

(D) $\frac{\sin 3 B}{\sin B}=\frac{3 \sin B-4 \sin ^3 B}{\sin B}=3-4 \sin ^2 B$
$=3-4+4 \cos ^2 B$
$=-1+\frac{4\left(a^2+c^2-b^2\right)^2}{4(a c)^2}$
$=-1+\frac{\left(\frac{a^2+c^2}{2}\right)^2}{(a c)^2} \ldots\left[\because 2 b^2=a^2+c^2\right]$
$=-1+\frac{\left( a ^2+ c ^2\right)^2}{4( ac )^2}$
$=\frac{\left( a ^2+ c ^2\right)^2-4 a ^2 c ^2}{4( ac )^2}$
$=\left(\frac{ c ^2- a ^2}{2 ac }\right)^2$

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