A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{6}\]
D) None of these
Correct Answer: B
Solution :
We have, \[\tan \,\alpha =\frac{m}{m+1}\] and \[\tan \,\beta =\frac{1}{2m+1}\] We know \[\tan \,(\alpha +\beta )=\frac{\tan \,\alpha +\tan \,\beta }{1-\tan \,\alpha \,\tan \,\beta }\] \[=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\frac{m}{(m+1)}\,\frac{1}{(2m+1)}}=\frac{2{{m}^{2}}+m+m+1}{2{{m}^{2}}+m+2m+1-m}\] \[=\frac{2{{m}^{2}}+2m+1}{2{{m}^{2}}+2m+1}=1\,\,\Rightarrow \,\,\tan \,(\alpha +\beta )=\tan \frac{\pi }{4}\] 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).
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