Trigonometric Identities Question 38
Question: If $ \tan \alpha ={{(1+{2^{-x}})}^{-1}}, $ $ \tan \beta ={{(1+{2^{x+1}})}^{-1}} $ , then $ \alpha +\beta $ equals
[AMU 2002]
Options:
A) $ \pi /6 $
B) $ \pi /4 $
C) $ \pi /3 $
D) $ \pi /2 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \tan ,(\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } $
Þ $ \tan (\alpha +\beta )=\frac{\frac{1}{1+\frac{1}{2^{x}}}+\frac{1}{1+{2^{x+1}}}}{1-\frac{1}{1+1/2^{x}}\frac{1}{1+{2^{x+1}}}} $
Þ $ \tan (\alpha +\beta )=\frac{2^{x}+{{2.2}^{x+x}}+2^{x}+1}{1+2^{x}+{{2.2}^{x}}+{{2.2}^{x+x}}-2^{x}} $
Þ $ \tan (\alpha +\beta )=1 $
$ \Rightarrow \alpha +\beta =\frac{\pi }{4} $ .
BETA
