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

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(C) Given: $\sec 4	heta - \sec 2	heta = 2$ $\Rightarrow \frac{1}{\cos 4	heta} - \frac{1}{\cos 2	heta} = 2$ $\Rightarrow \frac{\cos 2	heta - \cos

Vedclass · Accepted Answer

$n\pi + \frac{\pi}{2}$ or $\frac{n\pi}{5} + \frac{\pi}{10}$

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(C) Given: $\sec 4	heta - \sec 2	heta = 2$ $\Rightarrow \frac{1}{\cos 4	heta} - \frac{1}{\cos 2	heta} = 2$ $\Rightarrow \frac{\cos 2	heta - \cos 4	heta}{\cos 4	heta \cos 2	heta} = 2$ $\Rightarrow \cos 2	heta - \cos 4	heta = 2 \cos 4	heta \cos 2	heta$ Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$: $\Rightarrow \cos 2	heta - \cos 4	heta = \cos 6	heta + \cos 2	heta$ $\Rightarrow - \cos 4	heta = \cos 6	heta$ $\Rightarrow \cos 6	heta + \cos 4	heta = 0$ Using the formula $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$: $\Rightarrow 2 \cos 5	heta \cos 	heta = 0$ Case $1$: $\cos 5	heta = 0$ $\Rightarrow 5	heta = (2n + 1)\frac{\pi}{2}$ $\Rightarrow 	heta = (2n + 1)\frac{\pi}{10}$ Case $2$: $\cos 	heta = 0 \Rightarrow 	heta = (2n + 1)\frac{\pi}{2} = n\pi + \frac{\pi}{2}$ Thus,the general solution is $	heta = n\pi + \frac{\pi}{2}$ or $	heta = \frac{n\pi}{5} + \frac{\pi}{10}$.

If $\sec 4\theta - \sec 2\theta = 2$,then the general value of $\theta$ is

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