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Prove that. sec(theeta)/sec(theeta)-1 + sec(theeta)/sec(theeta)+1 = 2 cosec*2( theeta) |
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Answer» Answer: We have secθ+tan θcosecθ =secθ[ secθ secθ+tan 3 θcosecθ
] =secθ[1+tan 3 θ⋅ sinθ cosθ
] =secθ[1+tan 3 θ×cotθ] = 1+tan θ
[1+tan 2 θ] =[1+tan 2 θ] 3/2
=[1−(1−a 2 )] 3/2 =(2−a 2 ) 3/2
[∵tan 2 θ=1−a 2 ] Step-by-step EXPLANATION: PLEASE mark me as a Brainlist |
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