If `cosectheta-sintheta=a^3, sectheta-costheta=b^3`, then prove that `a^2b^2(a^2+b^2)=1`

Explanation

Given : `cosec theta-sin theta=a^{3}` and `sec theta-cos theta=b^{3}`



`Rightarrow frac{1}{sin theta}-sin theta `
`Rightarrow frac{cos ^{2} theta}{sin theta}=a^{3}`





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