Sequence And Series Question 203
Question: If $ 1+\cos \alpha +{{\cos }^{2}}\alpha +…….,\infty =2-\sqrt{2,} $ then $ \alpha , $ $ (0<\alpha <\pi ) $ is
[Roorkee 2000; AMU 2005]
Options:
A) $ \pi /8 $
B) $ \pi /6 $
C) $ \pi /4 $
D) $ 3\pi /4 $
Show Answer
Answer:
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} $ .
BETA
