Given: cosec(theta) - sin(theta) = a^3 ...(1)
sec(theta) - cos(theta) = b^3 ...(2)

To prove: a^2 * b^2 = 1



We can start by manipulating equations (1) and (2) to eliminate trigonometric functions and simplify the expressions.

From equation (1), we can rewrite it as:
cosec(theta) = sin(theta) + a^3 ...(3)

Similarly, from equation (2), we can rewrite it as:
sec(theta) = cos(theta) + b^3 ...(4)

Now, we can square both sides of equations (3) and (4) to eliminate the square root and simplify the expressions:

(cosec(theta))^2 = (sin(theta) + a^3)^2 ...(5)
(sec(theta))^2 = (cos(theta) + b^3)^2 ...(6)

Using the trigonometric identity for square of cosec(theta) and sec(theta):
(cosec(theta))^2 = 1 + (cot(theta))^2 ...(7)
(sec(theta))^2 = 1 + (tan(theta))^2 ...(8)

Substituting equations (7) and (8) into equations (5) and (6), we get:

1 + (cot(theta))^2 = (sin(theta) + a^3)^2 ...(9)
1 + (tan(theta))^2 = (cos(theta) + b^3)^2 ...(10)

Expanding the square terms on the right-hand side of equations (9) and (10), we have:

1 + (cot(theta))^2 = sin^2(theta) + 2a^3sin(theta) + a^6 ...(11)
1 + (tan(theta))^2 = cos^2(theta) + 2b^3cos(theta) + b^6 ...(12)

Now, we can use the trigonometric identity for cot(theta) and tan(theta):
cot(theta) = 1/tan(theta) ...(13)

Substituting equation (13) into equation (11), we get:

1 + (1/(tan(theta)))^2 = sin^2(theta) + 2a^3sin(theta) + a^6 ...(14)

Similarly, substituting equation (13) into equation (12), we have:

1 + (tan(theta))^2 = cos^2(theta) + 2b^3cos(theta) + b^6 ...(15)

Now, we can simplify equations (14) and (15) further by using the trigonometric identity for sin^2(theta) and cos^2(theta):

1 + (1/(tan(theta)))^2 = 1 - cos^2(theta) + 2a^3sin(theta) + a^6 ...(16)
1 + (tan(theta))^2 = 1 - sin^2(theta) + 2b^3cos(theta) + b^6 ...(17)

Rewriting equations (16) and (17) as:

1 + (1/(tan(theta)))^2 - 1 + cos^2(theta) = 2a^3sin(theta) + a^6