If cos (alpha - beta ) = 1 and cos (alpha + b | vedclass

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(D) Given $\cos(\alpha - \beta) = 1$. Since $-\pi < \alpha, \beta < \pi$,the range for the difference is $-2\pi < \alpha - \beta < 2\pi$.For $\cos(\al

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$4$

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(D) Given $\cos(\alpha - \beta) = 1$. Since $-\pi For $\cos(\alpha - \beta) = 1$,the possible values for $\alpha - \beta$ are $0$.Thus,$\alpha - \beta = 0$,which implies $\alpha = \beta$.Substituting $\alpha = \beta$ into the second equation: $\cos(\alpha + \alpha) = \cos(2\alpha) = \frac{1}{e}$.Since $-\pi In the interval $(-2\pi, 2\pi)$,the equation $\cos(2\alpha) = \frac{1}{e}$ (where $0 Therefore,there are $4$ ordered pairs $(\alpha, \beta)$.

If $\cos (\alpha - \beta ) = 1$ and $\cos (\alpha + \beta ) = \frac{1}{e}$,where $-\pi < \alpha, \beta < \pi$,then the total number of ordered pairs $(\alpha, \beta)$ is:

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