A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{6}\]
D) \[\frac{3\pi }{2}\]
Correct Answer: B
Solution :
\[\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }\] \[\,=\frac{\frac{m}{m+1}+\frac{1}{2\,\,m+1}}{1-\left( \frac{m}{m+1} \right)\left( \frac{1}{2\,\,m+1} \right)}\] \[=\frac{2{{m}^{2}}m+m+1}{2{{m}^{2}}+2m+m+1-m}\] \[\,=\frac{2{{m}^{2}}+2m+1}{2{{m}^{2}}+2m+1}=1\] \[\therefore \,\] \[\alpha +\beta ={{\tan }^{-1}}(1)=\pi /4\]
You need to login to perform this action.
You will be redirected in
3 sec
