If sec 4theta - sec 2theta = 2, then theta = | vedclass

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(C) Given $\frac{1}{\cos 4	heta} - \frac{1}{\cos 2	heta} = 2$. Taking $LCM$: $\frac{\cos 2	heta - \cos 4	heta}{\cos 4	heta \cos 2	heta} = 2$. Us

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$n\pi + \frac{\pi}{2}, \frac{n\pi}{5} + \frac{\pi}{10}, n \in Z$

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(C) Given $\frac{1}{\cos 4	heta} - \frac{1}{\cos 2	heta} = 2$. Taking $LCM$: $\frac{\cos 2	heta - \cos 4	heta}{\cos 4	heta \cos 2	heta} = 2$. Using $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$: $\frac{2 \sin 3	heta \sin 	heta}{\cos 4	heta \cos 2	heta} = 2$. $\sin 3	heta \sin 	heta = \cos 4	heta \cos 2	heta$. Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $2 \cos A \cos B = \cos(A-B) + \cos(A+B)$: $\frac{1}{2}(\cos 2	heta - \cos 4	heta) = \frac{1}{2}(\cos 2	heta + \cos 6	heta)$. $-\cos 4	heta = \cos 6	heta \implies \cos 6	heta + \cos 4	heta = 0$. $2 \cos 5	heta \cos 	heta = 0$. Case $1$: $\cos 5	heta = 0 \implies 5	heta = (2n+1)\frac{\pi}{2} \implies 	heta = \frac{(2n+1)\pi}{10} = \frac{n\pi}{5} + \frac{\pi}{10}$. Case $2$: $\cos 	heta = 0 \implies 	heta = (2n+1)\frac{\pi}{2} = n\pi + \frac{\pi}{2}$. Thus, $	heta = n\pi + \frac{\pi}{2}, \frac{n\pi}{5} + \frac{\pi}{10}$.

If $\sec 4\theta - \sec 2\theta = 2$, then $\theta =$

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