Prove that tan-1/2=Π/4-1/2cos-(4/5)
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prove that,tan⁻¹(1/2) = π/4 - 1/2cos⁻¹(4/5)
cos⁻¹(4/5) = Ф
cosФ = 4/5
(1 - tan²Ф/2)/(1 + tan²Ф/2) = 4/5
5 - 5tan²Ф/2 = 4 + 4tan²Ф/2
tan²Ф/2 = 1/9 = 1/(3)²
tanФ/2 = 1/3
means, we have to prove, 1/2 = tan[π/4 - Ф/2]
RHS = tan[π/4 - Ф/2]
= {tanπ/4 - tan(Ф/2)}/{1 + tanπ/4.tan(Ф/2)}
= { 1 - tanФ/2 }/{1 + tanФ/2}
= {1 - 1/3}/{1 + 1/3}
= {3 - 1}/{3 + 1}
= 2/4 = 1/2 = LHS
cos⁻¹(4/5) = Ф
cosФ = 4/5
(1 - tan²Ф/2)/(1 + tan²Ф/2) = 4/5
5 - 5tan²Ф/2 = 4 + 4tan²Ф/2
tan²Ф/2 = 1/9 = 1/(3)²
tanФ/2 = 1/3
means, we have to prove, 1/2 = tan[π/4 - Ф/2]
RHS = tan[π/4 - Ф/2]
= {tanπ/4 - tan(Ф/2)}/{1 + tanπ/4.tan(Ф/2)}
= { 1 - tanФ/2 }/{1 + tanФ/2}
= {1 - 1/3}/{1 + 1/3}
= {3 - 1}/{3 + 1}
= 2/4 = 1/2 = LHS
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