Q67 / 100 · BITSAT 2022 · Mathematics
If in a $\triangle A B C, 2 b^2=a^2+c^2$, then $\frac{\sin 3B}{\sin B}$ is equal to
- A$\frac{c^2-a^2}{2 c a}$
- B$\frac{c^2-a^2}{c a}$
- C$\frac{\left(c^2-a^2\right)^2}{(c a)^2}$
- Correct answerD$\left[\frac{c^2-a^2}{2 c a}\right]^2$
Solution
$\frac{\sin 3 B}{\sin B}=\frac{3 \sin B-4 \sin ^3 B}{\sin B}$
$=3-4 \sin ^2 B$
$=3-4\left(1-\cos ^2 B\right)$
$=-1+\frac{4\left(a^2+c^2-b^2\right)^2}{4(a c)^2}$
$=-1+\frac{\left(\frac{a^2+c^2}{2}\right)^2}{(a c)^2}$
$=-1+\frac{\left(a^2+c^2\right)^2}{4(a c)^2}$
$=(\frac{c^2-a^2}{2 a c})^2$
Hence, the correct answer is option: 'D'.