| If \[a\sin \theta +b\,\cos \theta =c,\] then \[a\cos \theta -b\sin \theta \] is equal to |
A) \[\pm \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]
B) \[\pm \sqrt{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}\]
C) \[\pm \sqrt{a+b-c}\]
D) \[\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}\]
Correct Answer: D
Solution :
| \[a\,\,\sin \theta +b\,\cos \theta =c\] |
| \[\Rightarrow \]\[{{(a\,\sin \theta +b\,\cos \theta )}^{2}}={{c}^{2}}\] |
| \[\Rightarrow \]\[{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2\,ab\sin \theta \cos ={{c}^{2}}\] |
| \[\Rightarrow \]\[{{a}^{2}}(1-{{\cos }^{2}}\theta )+{{b}^{2}}(1-{{\sin }^{2}}\theta )+2\,ab\sin \theta \cos \theta ={{c}^{2}}\] |
| \[\Rightarrow \]\[{{a}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}-{{b}^{2}}{{\sin }^{2}}\theta +2\,ab\sin \theta \cos \theta ={{c}^{2}}\] |
| \[\Rightarrow \]\[{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2\,ab\sin \theta \cos \theta \] |
| \[={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\] |
| \[\Rightarrow \] \[{{(a\,\cos \theta -b\,\sin \theta )}^{2}}={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\] |
| \[\Rightarrow \] \[a\,\cos \theta -b\sin \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] |
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