Trigonometric Identities Question 358
Question: If $ \tan \alpha =\frac{m}{m+1} $ and $ \tan \beta =\frac{1}{2m+1} $ , then $ \alpha +\beta = $
[IIT 1978; EAMCET 1992; Roorkee 1998; JMI EEE 2001]
Options:
A) $ \frac{\pi }{3} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{6} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have, $ \tan ,\alpha =\frac{m}{m+1} $ and $ \tan ,\beta =\frac{1}{2m+1} $ We know $ \tan ,(\alpha +\beta )=\frac{\tan ,\alpha +\tan ,\beta }{1-\tan ,\alpha ,\tan ,\beta } $ $ =\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\frac{m}{(m+1)},\frac{1}{(2m+1)}}=\frac{2m^{2}+m+m+1}{2m^{2}+m+2m+1-m} $ $ =\frac{2m^{2}+2m+1}{2m^{2}+2m+1}=1\Rightarrow \tan ,(\alpha +\beta )=\tan \frac{\pi }{4} $ Hence, $ \alpha +\beta =\frac{\pi }{4} $ . Trick : As $ \alpha +\beta $ is independent of m, therefore put $ m=1, $ then $ \tan ,\alpha =\frac{1}{2} $ and $ \tan ,\beta =\frac{1}{3} $ . Therefore, $ \tan ,(\alpha +\beta )=\frac{(1/2)+(1/3)}{1-(1/6)}=1. $ Hence $ \alpha +\beta =\frac{\pi }{4}. $ (Also check for other values of m).
BETA
