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Answer :
Given,
\( tan 2A = cot (A – 18°) \)
=>\( \ cot(90°–2A) = cot (90° – 18) \) [ \(\because tan \theta \ = \ cot (90° – \theta ) \) ]
=> \( 90° - 2A \ = \ A – 18° \) [ \(\because (90° – 2A) and (A – 18°) are both acute angle ]
=> \( 2A + A \ = \ 18° + 90° \)
=> \( 3A \ = \ 108° \)
=> \( A \ = \ 36° \)