If cos theta neq 0,and sec theta - 1 = (sqrt{ | vedclass

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(B) Given $\sec 	heta - 1 = (\sqrt{2} - 1) 	an 	heta$ with $\cos 	heta 
eq 0$.$\Rightarrow \frac{1 - \cos 	heta}{\cos 	heta} = (\sqrt{2} - 1) \

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$2 n \pi + \frac{\pi}{4} 	ext{ or } 2 n \pi, n \in Z$

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(B) Given $\sec 	heta - 1 = (\sqrt{2} - 1) 	an 	heta$ with $\cos 	heta 
eq 0$.$\Rightarrow \frac{1 - \cos 	heta}{\cos 	heta} = (\sqrt{2} - 1) \frac{\sin 	heta}{\cos 	heta}$Since $\cos 	heta 
eq 0$,we have $1 - \cos 	heta = (\sqrt{2} - 1) \sin 	heta$.Using half-angle identities $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$:$2 \sin^2 \frac{	heta}{2} = (\sqrt{2} - 1) 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$$\Rightarrow 2 \sin \frac{	heta}{2} [\sin \frac{	heta}{2} - (\sqrt{2} - 1) \cos \frac{	heta}{2}] = 0$This implies either $\sin \frac{	heta}{2} = 0$ or $	an \frac{	heta}{2} = \sqrt{2} - 1$.Case $1$: $\sin \frac{	heta}{2} = 0$ $\Rightarrow \frac{	heta}{2} = n \pi$ $\Rightarrow 	heta = 2 n \pi, n \in Z$.Case $2$: $	an \frac{	heta}{2} = \sqrt{2} - 1$. Since $	an \frac{\pi}{8} = \sqrt{2} - 1$,we have $\frac{	heta}{2} = n \pi + \frac{\pi}{8} \Rightarrow 	heta = 2 n \pi + \frac{\pi}{4}, n \in Z$.Thus,$	heta = 2 n \pi + \frac{\pi}{4} 	ext{ or } 2 n \pi, n \in Z$.

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

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