| If \[a\sin \theta +b\cos \theta =c,\]then the value of \[a\cos \theta -b\sin \theta ,\]is [SSC (CGL) 2013] |
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: D
Solution :
| Given, \[a\sin \theta +b\cos \theta =c\] |
| On squaring both sides, we get |
| \[{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2ab\sin \theta \cos \theta ={{c}^{2}}\] |
| \[[\because {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab]\] |
| \[\Rightarrow \]\[{{a}^{2}}(1-{{\cos }^{2}}\theta )+{{b}^{2}}\,\,(1-{{\sin }^{2}}\theta )+2ab\sin \theta \cos \theta ={{c}^{2}}\] |
| \[\Rightarrow \]\[{{a}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}-\,\,{{b}^{2}}{{\sin }^{2}}\theta +2ab\sin \theta \cos \theta ={{c}^{2}}\] |
| On rearranging, we get |
| \[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\sin \theta \cos \theta \]\[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}={{(a\cos \theta -b\sin \theta )}^{2}}\] |
| \[\Rightarrow \]\[a\cos \theta -b\sin \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}\] |
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