if cosecθ - sinθ = m and secθ - cosθ = n , prove that (m²n)^2/3 + (mn²)^2/3 = 1

Dear Student,
This question cannot be solved by n = secA -  tanA 
Please find below the solution to the asked query:
$$\begin{gathered} \mathrm{cosec} \mathrm{\theta }-\sin \mathrm{\theta } = \mathrm{m} \\ \Rightarrow \frac{1}{\sin \mathrm{\theta }}-\sin \mathrm{\theta } = \mathrm{m} \\ \Rightarrow \frac{1-{\sin }^2\mathrm{\theta }}{\sin \mathrm{\theta }} = \mathrm{m} \\ \Rightarrow \mathrm{m} = \frac{{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }} ...\left(\mathrm{i}\right) \\ \sec \mathrm{\theta }-\cos \mathrm{\theta } = \mathrm{n} \\ \Rightarrow \frac{1}{\cos \mathrm{\theta }}-\cos \mathrm{\theta } = \mathrm{n} \\ \Rightarrow \frac{1-{\cos }^2\mathrm{\theta }}{\cos \mathrm{\theta }} = \mathrm{n} \\ \Rightarrow \mathrm{n} = \frac{{\sin }^2\mathrm{\theta }}{\cos \mathrm{\theta }} ...\left(\mathrm{ii}\right) \\ \mathrm{So} \mathrm{we} \mathrm{have}; \\ {\left({\mathrm{m}}^2\mathrm{n}\right)}^{\frac{2}{3}}+{\left({\mathrm{mn}}^2\right)}^{\frac{2}{3}} \\ = {\left[{\left(\frac{{\cos }^2\mathrm{\theta }}{\mathrm{sin\theta }}\right)}^2\times \frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }}\right]}^{\frac{2}{3}}+{\left[\frac{{\cos }^2\mathrm{\theta }}{\mathrm{sin\theta }}\times {\left(\frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }}\right)}^2\right]}^{\frac{2}{3}} \\ = {\left[\frac{{\cos }^4\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}\times \frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }}\right]}^{\frac{2}{3}}+{\left[\frac{{\cos }^2\mathrm{\theta }}{\mathrm{sin\theta }}\times \frac{{\sin }^4\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}\right]}^{\frac{2}{3}} \\ = {\left({\cos }^3\mathrm{\theta }\right)}^{\frac{2}{3}}+{\left({\sin }^3\mathrm{\theta }\right)}^{\frac{2}{3}} \\ = {\left(\mathrm{cos\theta }\right)}^{3\times \frac{2}{3}}+{\left(\mathrm{sin\theta }\right)}^{3\times \frac{2}{3}} \\ = {\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta } \\ = 1 \\ \mathrm{hence} \mathrm{proved}. \end{gathered}$$

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