cot^2 alpha=1+a^2 then cosec alpha+cot^3 alpha sec alpha=

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$$\begin{gathered} {\cot }^2\mathrm{\alpha }=1+{\mathrm{a}}^2 \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{cosec}}^2\mathrm{\alpha }=1+{\cot }^2\mathrm{\alpha }=1+1+{\mathrm{a}}^2=2+{\mathrm{a}}^2 \\ \Rightarrow \mathrm{cosec\alpha }=\sqrt{2+{\mathrm{a}}^2}={\left(2+{\mathrm{a}}^2\right)}^{\frac{1}{2}}...\left(\mathrm{i}\right) \\ \mathrm{Now} \\ \mathrm{cosec\alpha }+{\cot }^3\mathrm{\alpha } \mathrm{sec\alpha } \\ =\mathrm{cosec\alpha }+\frac{{\cos }^3\mathrm{\alpha }}{{\sin }^3\mathrm{\alpha }} \frac{1}{\mathrm{cos\alpha }} \\ =\mathrm{cosec\alpha }+\frac{{\cos }^2\mathrm{\alpha }}{{\sin }^2\mathrm{\alpha }.\sin \mathrm{\alpha }} \\ =\mathrm{cosec\alpha }+{\cot }^2\mathrm{\alpha }.\mathrm{cosec\alpha } \\ =\mathrm{cosec\alpha }\left(1+{\cot }^2\mathrm{\alpha }\right) \\ =\mathrm{cosec\alpha }.{\mathrm{cosec}}^2\mathrm{\alpha } \\ ={\mathrm{cosec}}^3\mathrm{\alpha } \\ ={\left(2+{\mathrm{a}}^2\right)}^{\frac{3}{2}} .....\mathrm{Using} \left(\mathrm{i}\right) \\ \mathbf{\Rightarrow }\mathbf{cosec\alpha }\mathbf{+}{\mathbf{cot}}^{\mathbf{3}}\mathbf{\alpha }\mathbf{ }\mathbf{sec\alpha }\mathbf{=}{\left(\mathbf{2}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}} \end{gathered}$$

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