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) \[{{\left( \frac{{{c}^{2}}-{{a}^{2}}}{2ca} \right)}^{2}}\]
Correct Answer: D
Solution :
\[\frac{\sin 3B}{\sin B}=\frac{3\sin B-4{{\sin }^{3}}B}{\sin B}=3-4{{\sin }^{2}}B\] \[=3-4+4{{\cos }^{2}}B=-1+\frac{4{{({{a}^{2}}+{{c}^{2}}-{{b}^{2}})}^{2}}}{4{{(ac)}^{2}}}\] \[=-1+\frac{\left( \frac{{{a}^{2}}+{{c}^{2}}}{2} \right)}{{{(ac)}^{2}}}=-1+\frac{{{({{a}^{2}}+{{c}^{2}})}^{2}}}{4{{(ac)}^{2}}}\] \[=\frac{{{({{a}^{2}}+{{c}^{2}})}^{2}}-4{{a}^{2}}{{c}^{2}}}{4{{(ac)}^{2}}}={{\left( \frac{{{c}^{2}}-{{a}^{2}}}{2ac} \right)}^{2}}\].
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