Trigonometric Identities Question 197
Question: $ \frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha } $ is equal to
Options:
A) $ \sec ( \frac{\alpha }{2}-\frac{\pi }{8} ) $
B) $ \cos ( \frac{\pi }{8}-\frac{\alpha }{2} ) $
C) $ \tan ( \frac{\alpha }{2}-\frac{\pi }{8} ) $
D) $ \cot ( \frac{\alpha }{2}-\frac{\pi }{2} ) $
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Answer:
Correct Answer: C
Solution:
[c] $ \frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha } $ $ =\frac{\sqrt{2}-\sqrt{2}( \frac{1}{\sqrt{2}}\sin \alpha +\frac{1}{\sqrt{2}}\cos \alpha )}{\sqrt{2}( \frac{1}{\sqrt{2}}\sin \alpha -\frac{1}{\sqrt{2}}\cos \alpha )} $ $ =\frac{\sqrt{2}-\sqrt{2}\cos ( \alpha -\frac{\pi }{4} )}{\sqrt{2}\sin ( \alpha -\frac{\pi }{4} )} $ $ =\frac{\sqrt{2}(1-cos\theta )}{\sqrt{2}\sin \theta }, $ where $ \theta =\alpha -\frac{\pi }{4} $ $ \frac{2{{\sin }^{2}}(\theta /2)}{2\sin (\theta /2)cos(\theta /2)} $ $ =\tan \frac{\theta }{2}=\tan ( \frac{\alpha }{2}-\frac{\pi }{8} ) $
BETA
