A) \[\pi /8\]
B) \[\pi /6\]
C) \[\pi /4\]
D) \[3\pi /4\]
Correct Answer: D
Solution :
\[1-\cos \alpha =\frac{1}{2-\sqrt{2}}=1+\frac{1}{\sqrt{2}}\] Þ\[\cos \alpha =-\frac{1}{\sqrt{2}}=\cos \frac{3\pi }{4}\] Þ \[\alpha =\frac{3\pi }{4}\].
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