If in \[\Delta \,ABC\], \[2{{b}^{2}}={{a}^{2}}+{{c}^{2}},\]then \[\frac{\sin 3B}{\sin B}=\] [UPSEAT 1999]

\[\frac{{{c}^{2}}-{{a}^{2}}}{2ca}\]

\[\frac{{{c}^{2}}-{{a}^{2}}}{ca}\]

\[{{\left( \frac{{{c}^{2}}-{{a}^{2}}}{ca} \right)}^{2}}\]

\[{{\left( \frac{{{c}^{2}}-{{a}^{2}}}{2ca} \right)}^{2}}\]

If in \[\Delta \,ABC\], \[2{{b}^{2}}={{a}^{2}}+{{c}^{2}},\]then \[\fra

1.	If in \[\Delta \,ABC\], \[2{{b}^{2}}={{a}^{2}}+{{c}^{2}},\]then \[\frac{\sin 3B}{\sin B}=\] [UPSEAT 1999]
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» E.

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If in \[\Delta \,ABC\], \[2{{b}^{2}}={{a}^{2}}+{{c}^{2}},\]then \[\frac{\sin 3B}{\sin B}=\] [UPSEAT 1999]

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