Question

If in $\Delta ABC,$ $2b^2=a^2+c^2$, 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

The correct option is D ${\left(\frac{c^2-a^2}{2ca}\right)}^2$

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