Question

If $\cos \theta \neq 0$ and $\sec \theta - 1 = ( \sqrt{2} - 1 ) \tan \theta $ then $\theta $ =

• A. $n \pi + \frac{\pi}{8} , n \in Z $
• B. $2n\pi + \frac{\pi}{4} \; (or) \; 2n \pi , n \in Z $
• C. $2 n \pi + \frac{\pi}{8} , n \in Z $
• D. $2n\pi - \frac{\pi}{4} \; (or) \; 2n \pi , n \in Z $

Answer 1

The Correct Option is B

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