# If cosec theta-sin theta=a cube and sec theta -cos theta=b cube, prove that a square b square (a square+b square)=1

Monoxdifly
28 Points
4 years ago
cosec theta-sin theta=a cube
(1 / sin theta) – sin theta = a cube
(1 – sin theta squared) / sin theta = a cube
cos theta squared / sin theta = a cube
a cube = cos theta cot theta
a = (cos theta cot theta)1/3

sec theta -cos theta=b cube
(1 / cos theta) – cos theta = b cube
(1 – cos theta squared) / cos theta = b cube
sin theta squared / cos theta = b cube
b cube = sin theta tan theta
b = (sin theta tan theta)1/3

ab = (cos theta cot theta)1/3(sin theta tan theta)1/3 = (cos theta cot theta sin theta tan theta)1/3 = (sin theta cos theta)1/3
a3b3 = (ab)3 = sin theta cos theta

a square b square (a square+b square)=1
a4b2 + a2b4 = 1
a(a3b3/b) + b(a3b3/a) = 1
(a2 sin theta cos theta + b2 sin theta cos theta) / ab = 1
(a2 + b2)(sin theta cos theta) / (sin theta cos theta)1/3 = 1
((cos theta cot theta)1/3)2 + (sin theta tan theta)1/3)2) (sin theta cos theta)2/3 = 1
((cos theta cot theta)2/3 + (sin theta tan theta)2/3) (sin theta cos theta)2/3 = 1
(cos2theta cot theta sin theta)2/3 + (sin2theta tan theta cos theta)2/3 = 1
(cos2theta cos theta)2/3 + (sin2theta sin theta)2/3 = 1
(cos3theta)2/3 + (sin3theta)2/3 = 1
cos theta squared + sin theta squared = 1
1 = 1

Q.E.D