If $\;$ $\text{cosec} \theta - \sin \theta = a^3$, $\;$ $\sec \theta - \cos \theta = b^3$,
then prove that $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$
Given: $\;$ $\text{cosec } \theta - \sin \theta = a^3$
i.e. $\;$ $\dfrac{1}{\sin \theta} - \sin \theta = a^3$
i.e. $\;$ $a^3 = \dfrac{1 - \sin^2 \theta}{\sin \theta}$
i.e. $\;$ $a^3 = \dfrac{\cos^2 \theta}{\sin \theta}$ $\;\;\; \cdots \; (1)$
and $\;$ $\sec \theta - \cos \theta = b^3$
i.e. $\;$ $\dfrac{1}{\cos \theta} - \cos \theta = b^3$
i.e. $\;$ $b^3 = \dfrac{1 - \cos^2 \theta}{\cos \theta}$
i.e. $\;$ $b^3 = \dfrac{\sin^2 \theta}{\cos \theta}$ $\;\;\; \cdots \; (2)$
Dividing equations $(1)$ and $(2)$ gives
$\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \div \dfrac{\cos^2 \theta}{\sin \theta}$
i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \times \dfrac{\sin \theta}{\cos^2 \theta}$
i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^3 \theta}{\cos^3 \theta}$
i.e. $\;$ $\dfrac{b^3}{a^3} = \tan^3 \theta$
i.e. $\;$ $\dfrac{b}{a} = \tan \theta$
$\implies$ $\sin \theta = \dfrac{b}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (3)$
and $\;$ $\cos \theta = \dfrac{a}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (4)$
Substituting the values of $\sin \theta$ and $\cos \theta$ from equations $(3)$ and $(4)$ in equation $(1)$ gives
$a^3 = \dfrac{a^2}{a^2 + b^2} \div \dfrac{b}{\sqrt{a^2 + b^2}}$
i.e. $\;$ $a^3 = \dfrac{a^2}{a^2 + b^2} \times \dfrac{\sqrt{a^2 + b^2}}{b}$
i.e. $\;$ $a \; b = \dfrac{1}{\sqrt{a^2 + b^2}}$
i.e. $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$
Hence proved