If operatorname{cosec} theta - sin theta = m | vedclass

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(B) Given $m = \operatorname{cosec} 	heta - \sin 	heta = \frac{1}{\sin 	heta} - \sin 	heta = \frac{1 - \sin^2 	heta}{\sin 	heta} = \frac{\cos^2

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$1$

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(B) Given $m = \operatorname{cosec} 	heta - \sin 	heta = \frac{1}{\sin 	heta} - \sin 	heta = \frac{1 - \sin^2 	heta}{\sin 	heta} = \frac{\cos^2 	heta}{\sin 	heta}$.Given $n = \sec 	heta - \cos 	heta = \frac{1}{\cos 	heta} - \cos 	heta = \frac{1 - \cos^2 	heta}{\cos 	heta} = \frac{\sin^2 	heta}{\cos 	heta}$.Now,consider the expression $(m^2 n)^{2/3} + (m n^2)^{2/3}$.Substitute the values of $m$ and $n$:$(m^2 n)^{2/3} = \left( \frac{\cos^4 	heta}{\sin^2 	heta} \cdot \frac{\sin^2 	heta}{\cos 	heta} ight)^{2/3} = (\cos^3 	heta)^{2/3} = \cos^2 	heta$.Similarly,$(m n^2)^{2/3} = \left( \frac{\cos^2 	heta}{\sin 	heta} \cdot \frac{\sin^4 	heta}{\cos^2 	heta} ight)^{2/3} = (\sin^3 	heta)^{2/3} = \sin^2 	heta$.Adding these results:$\cos^2 	heta + \sin^2 	heta = 1$.

If $\operatorname{cosec} \theta - \sin \theta = m$ and $\sec \theta - \cos \theta = n$,then $(m^2 n)^{2/3} + (m n^2)^{2/3} = $

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