Trigonometric Identities Question 90
Question: If $ \sin \theta +\sin \varphi =a $ and $ \cos \theta +\cos \varphi =b, $ then $ \tan \frac{\theta -\varphi }{2} $ is equal to
[MP PET 1993]
Options:
A) $ \sqrt{\frac{a^{2}+b^{2}}{4-a^{2}-b^{2}}} $
B) $ \sqrt{\frac{4-a^{2}-b^{2}}{a^{2}+b^{2}}} $
C) $ \sqrt{\frac{a^{2}+b^{2}}{4+a^{2}+b^{2}}} $
D) $ \sqrt{\frac{4+a^{2}+b^{2}}{a^{2}+b^{2}}} $
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Answer:
Correct Answer: B
Solution:
Given that $ \sin \theta +\sin \varphi =a $ ?..(i) and $ \cos \theta +\cos \varphi =b $ ?..(ii) Squaring, $ {{\sin }^{2}}\theta +{{\sin }^{2}}\varphi +2\sin \theta \sin \varphi =a^{2} $ and $ {{\cos }^{2}}\theta +{{\cos }^{2}}\varphi +2\cos \theta \cos \varphi =b^{2} $ Adding, 2+ 2 $ (\sin \theta \sin \varphi +\cos \theta \cos \varphi )=a^{2}+b^{2} $
Þ $ 2\cos (\theta -\varphi )=a^{2}+b^{2}-2 $
Þ $ \cos (\theta -\varphi )=\frac{a^{2}+b^{2}-2}{2} $
$ \Rightarrow \frac{1-{{\tan }^{2}}\frac{\theta -\varphi }{2}}{1+{{\tan }^{2}}\frac{\theta -\varphi }{2}}=\frac{a^{2}+b^{2}-2}{2} $
Þ $ (a^{2}+b^{2})+(a^{2}+b^{2}){{\tan }^{2}}\frac{\theta -\varphi }{2}-2-2{{\tan }^{2}}\frac{\theta -\varphi }{2} $ $ =2-2{{\tan }^{2}}\frac{\theta -\varphi }{2} $
Þ $ \frac{4-a^{2}-b^{2}}{a^{2}+b^{2}}={{\tan }^{2}}\frac{\theta -\varphi }{2} $
Þ $ \tan \frac{(\theta -\varphi )}{2}=\sqrt{\frac{4-a^{2}-b^{2}}{a^{2}+b^{2}}} $ Trick: Put $ \theta =\frac{\pi }{2},\varphi =0^{o} $ , then $ a=1=b $ \ $ \tan \frac{\theta -\varphi }{2}=1 $ , which is given by (a) and (b). Again putting $ \theta =\frac{\pi }{4}=\varphi $ , we get $ \tan \frac{\theta -\varphi }{2}=0 $ , which is given by (b).
BETA
