A) \[a<b<c\]
B) \[a=b=c\]
C) \[{{c}^{2}}\le {{a}^{2}}+{{b}^{2}}\]
D) \[{{c}^{2}}<{{a}^{2}}-{{b}^{2}}\]
E) for all real values of a, b and c
Correct Answer: C
Solution :
The given equation is \[a\text{ }cos\theta +b\text{ }sin\theta =c\] Since, \[\sqrt{{{a}^{2}}-{{b}^{2}}}\le a\cos \theta +b\sin \theta \le \sqrt{{{a}^{2}}+{{b}^{2}}}\] \[\Rightarrow \] \[{{c}^{2}}\le \sqrt{{{a}^{2}}+{{b}^{2}}}\]
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