Posted by zohera on .
if cotx+tanx=a and secxcosx=b then prove that (a^2b)^0.6667 (ab^2)^0.6667=1

math 
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 = (seccos)^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