A) 0
B) \[\frac{1}{\sqrt{3}}\]
C) 1
D) \[\infty \]
Correct Answer: D
Solution :
Given, \[\tan \,A=\frac{1}{\sqrt{3}}=\tan 30{}^\circ \Rightarrow A=30{}^\circ \] \[\left[ \because \,\,\tan \,30{}^\circ =\frac{1}{\sqrt{3}} \right]\] And \[\tan \,B=\sqrt{3}=\tan \,60{}^\circ \Rightarrow B=60{}^\circ \] \[\left[ \because \,\,\tan \,60{}^\circ =\sqrt{3} \right]\] Now, \[A+B=30{}^\circ =60{}^\circ =90{}^\circ \] \[\therefore \,\,\,\tan \left( A+B \right)=\tan \,90{}^\circ =\infty \]
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