Question
if cotx+tanx=a and secx-cosx=b then prove that (a^2b)^0.6667 -(ab^2)^0.6667=1
Answers
Steve
a^2b^(2/3) - (ab^2)^(2/3)
(a^2b)^(1/3) - (ab^2)^(1/3)(a^2b)^(1/3) + (ab^2)^(1/3)
ignoring the x's for a bit,
a^2 = (cot+tan)^2 = cot^2 + 2 + tan^2 = csc^2 + sec^2
a^2b = sec^3
b^2 = (sec-cos)^2 = sec^2 - 2 + cos^2
ab^2 = tan^3
(a2b)^(2/3) = sec^2
(ab^2)^(2/3) = tan^2
sec^2 - tan^2 = 1
(a^2b)^(1/3) - (ab^2)^(1/3)(a^2b)^(1/3) + (ab^2)^(1/3)
ignoring the x's for a bit,
a^2 = (cot+tan)^2 = cot^2 + 2 + tan^2 = csc^2 + sec^2
a^2b = sec^3
b^2 = (sec-cos)^2 = sec^2 - 2 + cos^2
ab^2 = tan^3
(a2b)^(2/3) = sec^2
(ab^2)^(2/3) = tan^2
sec^2 - tan^2 = 1