Question

$\mathrm{If}\sec \theta +\tan \theta =x,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\sec \theta .$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \sec \theta +\tan \theta =x.....\left(\mathrm{i}\right) \\ \Rightarrow \frac{\sec \theta +\tan \theta }{1}\times \frac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta }=x \\ \Rightarrow \frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }=x \\ \Rightarrow \frac{1}{\sec \theta -\tan \theta }=\frac{x}{1} \\ \Rightarrow \sec \theta -\tan \theta =\frac{1}{x}.....\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get} \\ 2\sec \theta =x+\frac{1}{x} \\ \Rightarrow 2\sec \theta =\frac{x^2+1}{x} \\ \therefore \sec \theta =\frac{x^2+1}{2x} \end{gathered}$$