A) 0
B) 3
C) 1
D) 2
Correct Answer: D
Solution :
tan\[\left( {{\theta }_{1}}+{{\theta }_{2}} \right)\] \[=\sqrt{3}\]= tan \[60{}^\circ \] \[\Rightarrow \] \[{{\theta }_{1}}+{{\theta }_{2}}=60{}^\circ \] and sec \[({{\theta }_{1}}-{{\theta }_{2}})=\frac{2}{\sqrt{3}}\] = sec \[30{}^\circ \] \[\Rightarrow \]\[{{\theta }_{1}}+{{\theta }_{2}}=30{}^\circ \] \[\therefore \] \[{{\theta }_{1}}\] =\[45{}^\circ \] and \[{{\theta }_{2}}\]= \[15{}^\circ \] \[\therefore \] sin 2\[{{\theta }_{1}}\] + tan 3\[{{\theta }_{2}}\] = sin \[90{}^\circ \] + tan \[45{}^\circ \] =1+1=2
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