Correct option is (B) 36°
Given that,
\(\tan 2 A=\cot \left(A-18^{\circ}\right)\)
\(\cot \left(90^{\circ}-2 A\right) =\cot\left(A-18^{\circ}\right) \quad\left[\because \tan A=\cot\left(90^{\circ}-A\right)\right]\)
\(90^{\circ}-2 A=A-18^{\circ}\)
\(-2 A-A=-18^{\circ}-90^{\circ}\)
\(-3 A =-108^{\circ}\)
\(A =\frac{108}{3}=36^{\circ}\)
