Answer:
The given equation \[(1+{{m}^{2}}){{x}^{2}}+2mcx+{{c}^{2}}-{{a}^{2}}=0\] has equal roots Here, \[A=1+{{m}^{2}},B=2mc,C={{c}^{2}}-{{a}^{2}}\] For equal roots, \[~D=0={{B}^{2}}-4AC\] \[\Rightarrow \] \[{{(2mc)}^{2}}-4(1+{{m}^{2}})({{c}^{2}}-{{a}^{2}})=0\] \[\Rightarrow \] \[4{{m}^{2}}{{c}^{2}}-4({{c}^{2}}-{{a}^{2}}+{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{2}})=0\] \[\Rightarrow \] \[{{m}^{2}}{{c}^{2}}-{{c}^{2}}+{{a}^{2}}-{{m}^{2}}{{c}^{2}}+{{m}^{2}}{{a}^{2}}=0\] \[\Rightarrow \] \[-{{c}^{2}}+{{a}^{2}}(1+{{m}^{2}})=0\] \[\Rightarrow \] \[{{c}^{2}}={{a}^{2}}(1+{{m}^{2}})\] Hence Proved.
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