{"id":9185,"date":"2022-01-01T00:14:29","date_gmt":"2021-12-31T18:44:29","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9185"},"modified":"2022-01-01T00:41:22","modified_gmt":"2021-12-31T19:11:22","slug":"prove-that-cosec90-theta-sectheta","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-cosec90-theta-sectheta\/","title":{"rendered":"Prove that \\(Cosec(90 – \\theta)\\) = \\(Sec\\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
sec A = \\(sec\\theta\\) = \\(AC\\over AB\\) ……..(1)<\/p>\n
cosec C = \\(cosec(90 – \\theta)\\) = \\(AC\\over AB\\) ……..(2)<\/p>\n
From (1) and (2), we have<\/p>\n
\\(cosec(90 – \\theta)\\) = \\(sec\\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\\) sec A = \\(sec\\theta\\) = \\(AC\\over AB\\) ……..(1) cosec C = \\(cosec(90 – \\theta)\\) = \\(AC\\over AB\\) …<\/p>\n