If in $\triangle A B C$, with usual notations, $a^2, b^2, c^2$ are in A.P. then $\frac{\sin 3 B}{\sin B}=$

If in $\triangle A B C$, with usual notations, $a^2, b^2, c^2$ are in A.P. then $\frac{\sin 3 B}{\sin B}=$
  1. $\frac{a^2-c^2}{2 a c}$
  2. $\left(\frac{a^2-c^2}{2 a c}\right)^2$
  3. $\frac{\mathrm{a}^2-\mathrm{c}^2}{\mathrm{ac}}$
  4. $\left(\frac{a^2-c^2}{a c}\right)^2$

Solution

We have $2 b^2=a^2+c^2$ $\begin{aligned} & \frac{\sin 3 \mathrm{~B}}{\sin \mathrm{B}}=\frac{3 \sin \mathrm{B}-4 \sin ^2 \mathrm{~B}}{\sin \mathrm{B}} \\ & =3-4 \sin ^2 \mathrm{~B}=3-4\left(1-\cos ^2 \mathrm{~B}\right)=4 \cos ^2 \mathrm{~B}-1 \\ & =4\left[\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ac}}\right]^2-1=4\left[\frac{\mathrm{b}^2}{2 \mathrm{ac}}\right]-1 \\ & =\left(\frac{2 \mathrm{~b}^2}{2 \mathrm{ac}}\right)-1=\left(\frac{\mathrm{a}^2+\mathrm{c}^2}{2 \mathrm{ac}}\right)-1=\left(\frac{\mathrm{a}^2+\mathrm{c}^2}{2 \mathrm{ac}}+1\right)\left(\frac{\mathrm{a}^2+\mathrm{c}^2}{2 \mathrm{ac}}-1\right) \\ & =\frac{(\mathrm{a}+\mathrm{c})^2(\mathrm{a}-\mathrm{c})^2}{(2 \mathrm{ac})^2} \\ & =\left[\frac{(\mathrm{a}+\mathrm{c})(\mathrm{a}-\mathrm{c})}{2 \mathrm{ac}}\right]=\left(\frac{\mathrm{a}^2-\mathrm{c}^2}{2 \mathrm{ac}}\right)^2 \end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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