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

Solution :

Draw a right angled triangle ABC right angled at B.

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

sec A = \(sec\theta\) = \(AC\over AB\)                     ……..(1)

cosec C = \(cosec(90 – \theta)\) = \(AC\over AB\)             ……..(2)

From (1) and (2), we have

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