A) \[4{{b}^{2}}-{{a}^{2}}-6ac=0\]
B) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+abc=0\]
C) \[{{a}^{2}}+{{b}^{2}}+2abc-{{c}^{2}}=0\]
D) \[{{a}^{2}}+{{b}^{2}}-2ac=0\]
Correct Answer: A
Solution :
(a): \[\sin \theta +\cos \theta =\frac{2b}{a}\]_________(1) \[\sin \theta \times \cos \theta =\frac{3c}{a}\]__________(2) Squaring (1) and subtracting \[2\sin \theta \times \cos \theta \Rightarrow \] \[{{(sin\theta +cos\theta )}^{2}}-2\sin \theta \cos \theta ={{\sin }^{2}}\theta +{{\cos }^{2}}\theta \] \[=\frac{4{{b}^{2}}}{{{a}^{2}}}=\frac{6c}{a}\] \[\Rightarrow 1=\frac{4{{b}^{2}}-6ac}{{{a}^{2}}}\] \[\Rightarrow {{a}^{2}}=4{{b}^{2}}-6ac\] \[\Rightarrow 4{{b}^{2}}-{{a}^{2}}-6ac=0\]
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