Question: If $ a\sin \theta +b\cos \theta =c, $ then the value of $ a\cos \theta -b\sin \theta , $ is
Options:
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}} $
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Answer:
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}} $
BETA
