{"id":9184,"date":"2022-01-01T00:13:04","date_gmt":"2021-12-31T18:43:04","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9184"},"modified":"2022-01-01T00:40:55","modified_gmt":"2021-12-31T19:10:55","slug":"prove-that-sec90-theta-cosectheta","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-sec90-theta-cosectheta\/","title":{"rendered":"Prove that \\(Sec(90 – \\theta)\\) = \\(Cosec\\theta\\)."},"content":{"rendered":"
Draw a right angled triangle ABC right angled at B.<\/p>\n
Let \\(\\angle\\) A = \\(\\theta\\), then \\(\\angle\\) C = 90 – \\(\\theta\\)<\/p>\n
cosec A = \\(cosec\\theta\\) = \\(AC\\over BC\\) ……..(1)<\/p>\n
sec C = \\(sec(90 – \\theta)\\) = \\(AC\\over BC\\) ……..(2)<\/p>\n
From (1) and (2), we have<\/p>\n
\\(sec(90 – \\theta)\\) = \\(cosec\\theta\\)<\/p>\n","protected":false},"excerpt":{"rendered":"
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\\) …<\/p>\n