Question

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

• A. $\frac{(c^2–a^2)}{2ac}$
• B. $\frac{(c^2–a^2)}{ac}$
• C. ${\left(\frac{c^2–a^2}{ac}\right)}^2$
• D. ${\left[\frac{(c^2–a^2)}{2ac}\right]}^2$

Answer 1

The correct option is D

${\left[\frac{(c^2–a^2)}{2ac}\right]}^2$

Explanation for correct option:

Find the value of $\left(\frac{\sin 3B}{\sin B}\right)$:

Given,

$∆ABC,2b^2=a^2+c^2$

We know,

$\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{B}\mathbf{=}\mathbf{3}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{B}\mathbf{–}\mathbf{4}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{3}}\mathbf{}\mathit{B}$

Now.

$\begin{array}{rcl}\frac{\sin 3B}{\sin B}& =& 3–4{\sin }^{2}B\\ & =& 3–4(1–{\cos }^{2}B)\\ & =& 3–4+4{\cos }^{2}B\\ & =& 4{\cos }^{2}B–1\end{array}$

We know that $\cos B=\frac{a^2+c^2–b^2}{2ac}$

Substituting the values in above equation we get,

$\begin{array}{rcl}\frac{\sin 3B}{\sin B}& =& \frac{4{(a^2+c^2–b^2)}^2}{4a^2{c}^2}–1\\ & =& \frac{[{(a^2+c^2–b^2)}^2–a^2{c}^2]}{a^2{c}^2}\\ & =& \frac{[{(2b^2–b^2)}^2–a^2{c}^2}{a^2{c}^2}\\ & =& \frac{(b^4–a^2{c}^2)}{a^2{c}^2}\end{array}$ $2b^2=a^2+c^2$

Substitute $b^2=\frac{a^2+c^2}{2}$ in above equation and solve, we get

$$\begin{gathered} =\frac{(a^4+c^4–2a^2{c}^2)}{4a^2{c}^2} \\ =\frac{{(c^2–a^2)}^2}{{\left(2ac\right)}^2} \\ ={\left[\frac{(c^2–a^2)}{2ac}\right]}^2 \end{gathered}$$

Hence, the correct option is D.