Question :
(i) \(sin 18\over cos 72\)
(ii) \(tan 26\over cot 64\)
(iii) cos 48 – sin 42
(iv) cosec 31 – sec 59
Solution :
(i) \(sin 18\over cos 72\) = \(sin 18\over cos (90 – 18)\) = \(sin 18\over sin 18\) = 1
(ii) \(tan 26\over cot 64\) = \(tan 26\over cot (90 – 26)\) = \(tan 26\over tan 26\) = 1
(iii) cos 48 – sin 42 = cos(90 – 42) – sin 42 = sin 42 – sin 42 = 0
(iv) cosec 31 – sec 59 = cosec 31 – sec(90 – 31) = cosec 31 – cosec 31 = 0
