If tan β = n sin α cos α/1-n sin^2 α , prove that tan (α - β) = (1- n) tan α

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Aditya Gupta · Answer

thanks 4 askin ananya. t an (α - β) = [t anα - tanβ]/[1+tanαtanβ]....(1) but tan β = n sin α cos α/1-n sin^2α.......(2) , we see that sin α cos α=sin2α/2=tanα/[1+tan^2α] and sin^2α= tan^2 α/[1+tan^2 α] . now substitute these values in (2) and simplify, we get tanβ= ntanα/[1+tan^2α – ntan^2 α] substitute this value in (1), and simplify, we have t an (α - β) = t an (α)[1+tan^2α – ntan^2α – n]/[1+tan^2 α] factoring [1+tan^2 α – ntan^2α – n] as (1-n)(1+tan^2 α) we have =tanα (1-n)(1+tan^2α)/(1+tan^2 α) = (1- n) tan α hence proved

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If tan β = n sin α cos α/1-n sin^2 α , prove that tan (α - β) = (1- n) tan α

If tan β = n sin α cos α/1-n sin^2 α , prove that tan (α - β) = (1- n) tan α

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If tan β = n sin α cos α/1-n sin^2 α , prove that tan (α - β) = (1- n) tan α

If tan β = n sin α cos α/1-n sin^2 α , prove that tan (α - β) = (1- n) tan α

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