Trigonometric Identities Question 174
Question: If $ \tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }, $ then $ \sin \alpha +\cos \alpha $ and $ \sin \alpha -\cos \alpha $ must be equal to
[WB JEE 1971]
Options:
A) $ \sqrt{2}\cos \theta ,\sqrt{2}\sin \theta $
B) $ \sqrt{2}\sin \theta ,\sqrt{2}\cos \theta $
C) $ \sqrt{2}\sin \theta ,\sqrt{2}\sin \theta $
D) $ \sqrt{2},\cos \theta ,\sqrt{2},\cos \theta $
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Answer:
Correct Answer: A
Solution:
We have $ \tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha } $
$ \Rightarrow \tan \theta =\frac{\sin ( \alpha -\frac{\pi }{4} )}{\cos ( \alpha -\frac{\pi }{4} )}\Rightarrow \tan \theta =\tan ( \alpha -\frac{\pi }{4} ) $
$ \Rightarrow \theta =\alpha -\frac{\pi }{4}\Rightarrow \alpha =\theta +\frac{\pi }{4} $ Hence, $ \sin \alpha +\cos \alpha =\sin ( \theta +\frac{\pi }{4} )+\cos ( \theta +\frac{\pi }{4} ) $ $ =\sqrt{2}\cos \theta $ and $ \sin \alpha -\cos \alpha =\sin ( \theta +\frac{\pi }{4} )-\cos ( \theta +\frac{\pi }{4} ) $ $ =\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta $ $ =\frac{2}{\sqrt{2}}\sin \theta =\sqrt{2}\sin \theta $ .
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