Solution :
Draw a right angled triangle ABC right angled at B.
Let \(\angle\) A = \(\theta\), then \(\angle\) C = 90 – \(\theta\)
tan A = \(tan\theta\) = \(BC\over AB\) ……..(1)
cot C = \(cot(90 – \theta)\) = \(BC\over AB\) ……..(2)
From (1) and (2), we have
\(cot(90 – \theta)\) = \(tan\theta\)