Prove that \(Sec(90 – \theta)\) = \(Cosec\theta\).

Solution :

Draw a right angled triangle ABC right angled at B.

Let \(\angle\) A = \(\theta\),  then \(\angle\) C = 90 – \(\theta\)

cosec A = \(cosec\theta\) = \(AC\over BC\)                     ……..(1)

sec C = \(sec(90 – \theta)\) = \(AC\over BC\)             ……..(2)

From (1) and (2), we have

\(sec(90 – \theta)\) = \(cosec\theta\)