Prove that $Tan(90 – \theta)$ = $Cot\theta$.

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

$tan(90 – \theta)$ = $cot\theta$