Question: If $ a\sin \theta +b,\cos \theta =c, $ then $ a\cos \theta -b\sin \theta $ is equal to
Options:
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}} $
Show Answer
Answer:
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}} $
BETA
