Find the value of λ in 3−tan2π71−tan2π7=λcosπ7
The numerator looks similar to expansion of tan3x, so I tried this
3tanπ7−tan3π7tanπ7(1−tan2π7)=λcosπ7
(3tanπ7−tan3π7)(1−3tan2π7)tanπ7(1−tan2π7)(1−3tan2π7)=λcosπ7
tan3π7(1−3tan2π7)sinπ7(1−tan2π7)=λ
But I'm stuck here. Need help. Thanks in advance.
Answer
Let a=π/7 then we have 1+21−tan2a=λcosa
multiplying this with tana we get: tana+tan2a=λsina
so sinacos2a+sin2acosa=λsinacosacos2a
multiplying this with 4 we get 4sin3a=λsin4a
Sinnce sin3a=sin4a we get λ=4.
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