\[\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha }=\] [AMU 1999]

\[\cot \left( \frac{\alpha }{2}-\frac{\pi }{2} \right)\]

\[\sec \left( \frac{\alpha }{2}-\frac{\pi }{8} \right)\]

\[\cos \left( \frac{\pi }{8}-\frac{\alpha }{2} \right)\]

\[\tan \left( \frac{\alpha }{2}-\frac{\pi }{8} \right)\]

\[\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha }

1.	\[\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha }=\] [AMU 1999]
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.	\[\cot \left( \frac{\alpha }{2}-\frac{\pi }{2} \right)\]
Answer» D. \[\cot \left( \frac{\alpha }{2}-\frac{\pi }{2} \right)\]

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\[\frac{\sqrt{2}-\sin \alpha -\cos \alpha }{\sin \alpha -\cos \alpha }=\] [AMU 1999]

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