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 :
\[\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha }\] =\[\frac{\sqrt{2}-\sqrt{2}\left\{ \frac{1}{\sqrt{2}}\sin \alpha +\frac{1}{\sqrt{2}}\cos \alpha \right\}}{\sqrt{2}\left\{ \frac{1}{\sqrt{2}}\sin \alpha -\frac{1}{\sqrt{2}}\cos \alpha \right\}}\]=\[\frac{\sqrt{2}-\sqrt{2}\cos \left( \alpha -\frac{\pi }{4} \right)}{\sqrt{2}\sin \left( \alpha -\frac{\pi }{4} \right)}\] = \[\frac{\sqrt{2}\left\{ \,1-\cos \theta \right\}}{\sqrt{2}\sin \theta },\] 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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