How do you prove #(cos2A)/sinA + (sin2A)/cosA = csc A#?

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Answer 1 · 2018-07-12T13:27:44

Please refer to a Proof in the Explanation.

#(cos2A)/sinA+(sin2A)/cosA#,

#=(cos2AcosA+sin2AsinA)/(sinAcosA)#,

#=(cos(2A-A))/(sinAcosA)#,

#=cosA/(sinAcosA)#,

#=1/sinA#,

#=cscA#, as desired!

Answer 2 · 2018-07-12T18:32:22

Please refer to a Second Proof in Explanation.

#"Using "cos2A=1-2sin^2A, and, sin2A=2sinAcosA#,

#"The L.H.S."=(1-2sin^2A)/sinA+(2sinAcosA)/cosA#,

#=1/sinA-(2sin^2A)/sinA+2sinA#,

#=cscA-2sinA+2sinA#,

#=cscA#,

#="The R.H.S."#