see attachment and explain it

Given that cosec θ – sin θ
= b³
and sec θ – cos θ = a³
Solving the first equation we get,
1/ sin θ – sin θ = b³
This gives (1-sin²θ) / sin θ = b³
This yields cos²θ/ sin θ = b³
Similarly, solving the second equation on similar lines gives
sin²θ/ cos θ = a³
Now, considering L.H.S
a⁴b² + a²b⁴
= (sin²θ/ cos θ)^4/3 . (cos²θ/ sin θ)^2/3 + (sin²θ/ cos θ)^2/3 . (cos²θ/ sin θ)^4/3
On simplifying this we get,
= sin ²θ + cos²θ
= 1.
Hence, a⁴b² + a²b⁴ = 1.

Approved