Question

If (sec θ + tan θ) = m and (sec θ − tan θ) = n, show that mn = 1.

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \text{We have}(\sec \theta +\tan \theta )=m...\left(\text{i}\right)\\ \text{Again,}(\sec \theta -\tan \theta )=n...\left(\text{ii}\right)\\ \text{Now, multiplying}\left(\text{i}\right)\text{and}\left(\text{ii}\right),\mathrm{we}\mathrm{get}:\end{array} \\ \begin{array}{l}(\sec \theta +\tan \theta )\times (\sec \theta -\tan \theta )=mn\\ =>{\sec }^2\theta -{\tan }^2\theta =mn\\ =>1=mn\text{[}\because {\sec }^2\theta -{\tan }^2\theta =1]\\ \therefore mn=1\end{array} \end{gathered}$$