Trigonometric Identities Question 172
Question: If $ \tan \theta +\sin \theta =m $ and $ \tan \theta -\sin \theta =n, $ then
[IIT 1970]
Options:
A) $ m^{2}-n^{2}=4mn $
B) $ m^{2}+n^{2}=4mn $
C) $ m^{2}-n^{2}=m^{2}+n^{2} $
D) $ m^{2}-n^{2}=4\sqrt{mn} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ (m+n)=2,\tan \theta ,m-n=2,\sin \theta $
$ \therefore ,m^{2}-n^{2}=4,\tan \theta ,.,\sin \theta $ ?..(i) $ 4\sqrt{mn}=4\sqrt{{{\tan }^{2}}\theta -{{\sin }^{2}}\theta }=4,\sin \theta ,.,\tan \theta $ ?..(ii) From (i) and (ii), $ m^{2}-n^{2}=4\sqrt{mn} $ .
BETA
