A) \[\tan \theta =\frac{3}{4}\]
B) \[\tan \theta =\frac{4}{3}\]
C) \[\tan \theta =\frac{2m}{{{m}^{2}}+1}\]
D) None of these
Correct Answer: B
Solution :
Squaring the given relation and putting \[\tan \theta =t,\] \[{{(m+2)}^{2}}\,{{t}^{2}}+2(m+2)\,(2m-1)t+{{(2m-1)}^{2}}={{(2m+1)}^{2}}\,(1+{{t}^{2}})\] \[\Rightarrow \,3\,(1-{{m}^{2}})\,{{t}^{2}}+(4{{m}^{2}}+6m-4)\,t-8m=0\] \[\Rightarrow \,(3t-4)\,[(1-{{m}^{2}})\,t+2m]=0\], which is true if \[t=\tan \theta =\frac{4}{3}\] or \[\tan \theta =\frac{2m}{{{m}^{2}}-1}\].
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