Question

$\mathrm{If}5x=\sec \theta \mathrm{and}\frac{5}{x}=\tan \theta ,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}5\left(x^2-\frac{1}{x^2}\right).\left[\mathrm{CBSE}2010\right]$

Answer 1

$$\begin{gathered} 5\left(x^2-\frac{1}{x^2}\right) \\ =\frac{25}{5}\left(x^2-\frac{1}{x^2}\right) \\ =\frac{1}{5}\left(25x^2-\frac{25}{x^2}\right) \\ =\frac{1}{5}\left[{\left(5x\right)}^2-{\left(\frac{5}{x}\right)}^2\right] \\ =\frac{1}{5}\left[{\left(\sec \theta \right)}^2-{\left(\tan \theta \right)}^2\right] \\ =\frac{1}{5}\left({\sec }^2\theta -{\tan }^2\theta \right) \\ =\frac{1}{5}\left(1\right) \\ =\frac{1}{5} \end{gathered}$$