Question

if cotx+tanx=a and secx-cosx=b then prove that (a^2b)^0.6667 -(ab^2)^0.6667=1

Answer 1

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