Question

Solve $\sqrt{2}\sec \theta +\tan \theta =1$ $\theta \in 2n\pi -\frac{\pi }{m}$. Find m.

Answer 1

Given, $\sqrt{2}\sec \theta +\tan \theta =1$
$\Rightarrow \frac{\sqrt{2}}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=1$
$\Rightarrow \sin \theta -\cos \theta =-\sqrt{2}$
$\Rightarrow \frac{1}{\sqrt{2}}\sin \theta -\frac{1}{\sqrt{2}}\cos \theta =-1$
$\Rightarrow \cos (\theta +\frac{\pi }{4})=1$
$\Rightarrow \theta +\frac{\pi }{4}=2n\pi \Rightarrow \theta =2n\pi -\frac{\pi }{4}$