Question

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

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

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 $\frac{\theta }{2}=n\pi +\frac{\pi }{8},n\in Z$
$\Rightarrow \theta =2n\pi$
or $\theta =2n\pi +\frac{\pi }{4}$