A) \[\pm \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]
B) \[\pm \sqrt{{{a}^{2}}-{{b}^{2}}+{{c}^{2}}}\]
C) \[\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}\]
D) \[\pm \sqrt{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}}\]
Correct Answer: C
Solution :
Given, \[a\cos \theta -b\sin \theta =c\] On squaring, we get \[{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\cos \theta \sin \theta ={{c}^{2}}\]\[\Rightarrow \]\[{{a}^{2}}(1-{{\sin }^{2}}\theta )+{{b}^{2}}(1-{{\cos }^{2}}\theta )-2ab\sin \theta \cos \theta ={{c}^{2}}\] \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}={{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2ab\sin \theta \cos \theta \] \[\Rightarrow \]\[{{(a\sin \theta +b\cos \theta )}^{2}}={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\] \[\therefore \]\[a\sin \theta +b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}\]
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