A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) zero
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
Key idea: \[tan(\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\] Given \[\tan \alpha =\frac{m}{m+1},\tan \beta =\frac{1}{2m+1}\] ...(i) \[\therefore \] \[\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\] \[=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\frac{m}{m+1}.\frac{1}{2m+1}}\] [using Eq. (i)] \[=\frac{2{{m}^{2}}+m+m+1}{2{{m}^{2}}+m+2m+1-m}=\frac{2{{m}^{2}}+2m+1}{2{{m}^{2}}+2m+1}\] \[\therefore \] \[\tan (\alpha +\beta )=1\] \[\Rightarrow \] \[\alpha +\beta ={{\tan }^{-1}}1=\frac{\pi }{4}\] Alternative Method Given, \[\tan \alpha =\frac{m}{m+1},\tan \beta =\frac{1}{2m+1}\] \[\therefore \]\[\alpha ={{\tan }^{-1}}\left( \frac{m}{m+1} \right),\beta ={{\tan }^{-1}}\left( \frac{1}{2m+1} \right)\] \[\alpha +\beta ={{\tan }^{-1}}\left( \frac{m}{m+1} \right)+{{\tan }^{-1}}\left( \frac{1}{2m+1} \right)\] \[={{\tan }^{-1}}\left( \frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\frac{m}{m+1}.\frac{1}{2m+1}} \right)\] \[\left[ \because {{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right) \right]\] \[\therefore \]\[\alpha +\beta ={{\tan }^{-1}}\left( \frac{2{{m}^{2}}+2m+1}{2{{m}^{2}}+2m+1} \right)\] \[={{\tan }^{-1}}1=\frac{\pi }{4}\]
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