{"id":9204,"date":"2022-01-03T02:10:55","date_gmt":"2022-01-02T20:40:55","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9204"},"modified":"2022-01-03T02:13:11","modified_gmt":"2022-01-02T20:43:11","slug":"prove-that-1-cot2-theta-cosec2-theta","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-1-cot2-theta-cosec2-theta\/","title":{"rendered":"Prove that 1 + \\(cot^2 \\theta\\) = \\(cosec^2 \\theta\\)."},"content":{"rendered":"
In right angled triangle ABC,<\/p>\n
\\(cosec \\theta\\) = \\(AC\\over BC\\) \\(\\implies\\) \\(cosec^2 \\theta\\) = \\(AC^2\\over BC^2\\)<\/p>\n
\\(cot \\theta\\) = \\(AB\\over BC\\) \\(\\implies\\) \\(cot^2 \\theta\\) = \\(AB^2\\over BC^2\\)<\/p>\n
\\(\\implies\\) 1 + \\(cot^2 \\theta\\) = 1 + \\(AB^2\\over BC^2\\) = \\(BC^2 + AB^2\\over BC^2\\) = \\(AC^2\\over BC^2\\)<\/p>\n
[ By Pythagoras theorem, \\(AC^2\\) = \\(BC^2 + AB^2\\) ]<\/p>\n
\\(\\implies\\) 1 + \\(cot^2 \\theta\\) = \\(cosec^2 \\theta\\)<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : In right angled triangle ABC, \\(cosec \\theta\\) = \\(AC\\over BC\\) \\(\\implies\\) \\(cosec^2 \\theta\\) = \\(AC^2\\over BC^2\\) \\(cot \\theta\\) = \\(AB\\over BC\\) \\(\\implies\\) \\(cot^2 \\theta\\) = \\(AB^2\\over BC^2\\) \\(\\implies\\) 1 + \\(cot^2 \\theta\\) = 1 + \\(AB^2\\over BC^2\\) = \\(BC^2 + AB^2\\over BC^2\\) = \\(AC^2\\over BC^2\\) [ By Pythagoras theorem, \\(AC^2\\) = \\(BC^2 + …<\/p>\n