A) \[\sec \left( \frac{\alpha }{2}-\frac{\pi }{8} \right)\]
B) \[\cos \left( \frac{\pi }{8}-\frac{\alpha }{2} \right)\]
C) \[\tan \left( \frac{\alpha }{2}-\frac{\pi }{8} \right)\]
D) \[\cot \left( \frac{\alpha }{2}-\frac{\pi }{2} \right)\]
Correct Answer: C
Solution :
[c] \[\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha }\] \[=\frac{\sqrt{2}-\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin \alpha +\frac{1}{\sqrt{2}}\cos \alpha \right)}{\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin \alpha -\frac{1}{\sqrt{2}}\cos \alpha \right)}\] \[=\frac{\sqrt{2}-\sqrt{2}\cos \left( \alpha -\frac{\pi }{4} \right)}{\sqrt{2}\sin \left( \alpha -\frac{\pi }{4} \right)}\] \[=\frac{\sqrt{2}(1-cos\theta )}{\sqrt{2}\sin \theta },\]where \[\theta =\alpha -\frac{\pi }{4}\] \[\frac{2{{\sin }^{2}}(\theta /2)}{2\sin (\theta /2)cos(\theta /2)}\] \[=\tan \frac{\theta }{2}=\tan \left( \frac{\alpha }{2}-\frac{\pi }{8} \right)\]
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