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