Question

Solve:

${\sec }^2\theta =\sqrt{2}(1-{\tan }^2\theta )$

Answer 1

Given, ${\sec }^2\theta =\sqrt{2}(1-{\tan }^2\theta )$
$\Rightarrow \frac{1}{{\cos }^2\theta }=\sqrt{2}-\frac{\sqrt{2}{\sin }^2\theta }{{\cos }^2\theta }$

$\Rightarrow 1=\sqrt{2}{\cos }^2\theta -\sqrt{2}{\sin }^2\theta$

$\Rightarrow 1=\sqrt{2}\cos 2\theta \Rightarrow \cos 2\theta =\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}$
$\Rightarrow 2\theta =2n\pi \pm \frac{\pi }{4}\Rightarrow \theta =n\pi \pm \frac{\pi }{8}$