Question

If $\sec \theta-1=(\sqrt{2}-1) \tan \theta$, then $\theta=$

• A. $(2 n-1) \pi$
• B. $2 n \pi+\frac{\pi}{4}$
• C. $2 n \pi-\frac{\pi}{4}$
• D. $2 n \pi+\frac{\pi}{3}$

Answer 1

Answer: (B)

$\sec \theta-1=(\sqrt{2}-1) \tan \theta$
$\frac{1-\cos \theta}{\cos \theta}=\frac{(\sqrt{2}-1) \sin \theta}{\cos \theta}$
$\Rightarrow 2 \sin ^{2}(\theta / 2)=(\sqrt{2}-1) 2 \sin (\theta / 2) \cos (\theta / 2)$
$\Rightarrow \sin (\theta / 2)=0$
or $\tan (\theta / 2)=(\sqrt{2}-1)=\tan (\pi / 8)$
$\Rightarrow \theta / 2=n \pi$
or $\frac{\theta}{2}=n \pi+\frac{\pi}{8}, n \in Z$
$\Rightarrow \theta=2 n \pi$
or $\theta=2 n \pi+\frac{\pi}{4}$