Question

If (sec θ – tan θ) = $\sqrt{2}$ tan θ then prove that (sec θ + tan θ) = $\sqrt{2}$ sec θ.

Answer 1

$\sec \theta -\tan \theta =\sqrt{2}\tan \theta$
Squaring on both sides, we get
$$\begin{gathered} {\left(\sec \theta -\tan \theta \right)}^2={\left(\sqrt{2}\tan \theta \right)}^2 \\ \Rightarrow {\sec }^2\theta +{\tan }^2\theta -2\sec \theta \tan \theta =2{\tan }^2\theta \\ \Rightarrow {\sec }^2\theta -{\tan }^2\theta =2\sec \theta \tan \theta \\ \Rightarrow \left(\sec \theta -\tan \theta \right)\left(\sec \theta +\tan \theta \right)=2\sec \theta \tan \theta \left[a^2-b^2=\left(a-b\right)\left(a+b\right)\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow \sqrt{2}\tan \theta \left(\sec \theta +\tan \theta \right)=2\sec \theta \tan \theta \left(\sec \theta -\tan \theta =\sqrt{2}\tan \theta \right) \\ \Rightarrow \sec \theta +\tan \theta =\frac{2\sec \theta \tan \theta }{\sqrt{2}\tan \theta } \\ \Rightarrow \sec \theta +\tan \theta =\sqrt{2}\sec \theta \end{gathered}$$