Question

If ${\tan }^2\mathrm{\theta }=1-e^2,$ prove that $\sec \mathrm{\theta }+{\tan }^3\mathrm{\theta }\mathrm{cosec}\mathrm{\theta }=2{\left(2-e^2\right)}^{3/2}$.

Answer 1

$$\begin{gathered} {\tan }^2\theta =1-e^2 \\ \Rightarrow se{c}^2\theta -1=1-e^2 \\ \Rightarrow se{c}^2\theta =2-e^2 \\ \Rightarrow {\cos }^2\theta =\frac{1}{2-e^2}\left(1\right) \\ \mathrm{LHS}: \\ sec\theta +{\tan }^3\theta .\cos ec\theta \\ =\frac{1}{\cos \theta }+\frac{{\sin }^2\theta }{{\cos }^3\theta } \\ =\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^3\theta } \\ =\frac{1}{{\cos }^3\theta } \\ ={\left(\cos \theta \right)}^{-3} \\ ={\left(\frac{1}{2-e^2}\right)}^{-\frac{3}{2}}\left(\mathrm{From}\right(1\left)\right) \\ ={\left(2-e^2\right)}^{\frac{3}{2}} \end{gathered}$$

Hence proved.