Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $2b^2=a^2+c^2$, then $\frac{\sin 3B}{\sin B}$ is equal to?

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

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\frac{c^2-a^2}{2ca}$
C ${\left(\frac{c^2-a^2}{ca}\right)}^2$
D ${\left(\frac{c^2-a^2}{2ca}\right)}^2$

$\frac{\sin 3B}{\sin B}=\frac{3\sin B-4\sin {}^{3}B}{\sin B}=3-4\sin {}^{2}B$

Given, $2b^2=a^2+c^2$

We Know, $\cos B=\frac{a^2+c^2-b^2}{2ac}\Rightarrow \frac{2b^2-b^2}{2ac}\left[CosineRule\right]\cos B=\frac{b^2}{2ac}$

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

Hence, $\frac{c^2-a^2}{2ca}$, ${\left(\frac{c^2-a^2}{ca}\right)}^2$, ${\left(\frac{c^2-a^2}{2ca}\right)}^2$ are the correct answers.