If cosecA-sinA=x^3 and secA-cosA=y^3, then show that x^2y^2{x^2+y^2}=1

`csecA - sinA= x^3
`

`==> 1/sinA - sinA =
x^3`

`==> (1-sin^2 A)/sinA =
x^3`

`==> (cos^2 A)/sinA = x^3
`

`==> x=
((cos^2A)/sinA)^(1/3)...............(1)`

`secA - cosA =
y^3`

`==> 1/cosA - cosA =
y^3`

`==> (1-cos^2 A)/cosA =
y^3`

`==> (sin^2 A)/cosA =
y^3`

`==> y= ((sin^2
A)/cosA)^(1/3)......................(2)`

Now we will
substitute into the equations:

`(x^2 y^2)(x^2+y^2) =
1`

`((cos^2 A)/sinA)^(2/3) ((sin^2 A)/cosA)^(2/3) (((cos^2
A)/sinA)^(2/3) + ((sin^2 A)/cosA)^(2/3)) = 1`

`((cosA
sinA)^(4/3) )/ (sinA cosA)^(2/3) (((cos^(4/3) A)(cos^ (2/3) A)+(sin^(4/3) A) (sin^(2/3)
A))/(sinA cosA)^(2/3))= 1`

`==> (cosA sinA)^(2/3)(((
cos^2 A) + (sin^2 A))/ (sinA cosA)^(2/3))= 1`

Reduce
similar terms.

==> `cos^2 A + sin^2 A =
1`

`==> 1 = 1`

``Then,
we have proved that if  `secA -cosx = y^3 `   and  `csecA-sinA= x^3 `  Then, `(x^2 y^2)
(x^2 +y^2) =
1`

``

``

``