Trigonometric Identities Question 175
Question: If $ a\cos \theta +b\sin \theta =m $ and $ a\sin \theta -b\cos \theta =n, $ then $ a^{2}+b^{2}= $
Options:
A) $ m+n $
B) $ m^{2}-n^{2} $
C) $ m^{2}+n^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ a,\cos \theta +b,\sin \theta =m $ and $ a,\sin ,\theta -b,\cos \theta =n. $ Squaring and adding, we get $ {{(a\cos \theta +b,\sin \theta )}^{2}}+{{(a,\sin \theta -b,\cos \theta )}^{2}}=m^{2}+n^{2} $
$ \Rightarrow a^{2}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+b^{2}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta ) $ $ +2ab,(\cos \theta ,\sin \theta -\sin \theta ,\cos \theta )=m^{2}+n^{2} $ Hence, $ a^{2}+b^{2}=m^{2}+n^{2}. $ Trick: Here we can guess that the value of $ a^{2}+b^{2} $ is independent of q, so put any suitable value of q i.e. $ \frac{\pi }{2}, $ so that $ b=m $ and $ a=n. $ Hence $ a^{2}+b^{2}=m^{2}+n^{2}. $ (Also check for other value of q).
BETA
