If `cos e ctheta-sintheta=a^3,sectheta-costheta=b^3,` Prove that : `a^2b^2(a^2+b^2)=1`

Explanation

`cosec theta-sin theta=a^{3}`

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

`text { (i.e) } frac{1-sin ^{2} theta}{sin theta}=a^{3}`

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

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

`text { Similary, } sec theta-cos theta=b^{3}`

`Rightarrow frac{1}{cos theta}-cos theta=b^{3}`

`frac{1-cos ^{2} theta}{cos theta}=b^{3}`

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