If frac{1 - cos x}{cos x(1 + cos x)} = frac{s | vedclass

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(C) Given the equation: $\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{\sin \alpha}{\cos x} - \frac{2}{1 + \cos x}$ Multiply both sides by $\cos x(1 +

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$\frac{\pi}{2}$

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(C) Given the equation: $\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{\sin \alpha}{\cos x} - \frac{2}{1 + \cos x}$ Multiply both sides by $\cos x(1 + \cos x)$: $1 - \cos x = \sin \alpha (1 + \cos x) - 2 \cos x$ $1 - \cos x = \sin \alpha + \sin \alpha \cos x - 2 \cos x$ $1 - \cos x = \sin \alpha + \cos x(\sin \alpha - 2)$ Comparing the constant terms and the coefficients of $\cos x$ on both sides: Constant term: $1 = \sin \alpha$ Coefficient of $\cos x$: $-1 = \sin \alpha - 2$ From the constant term,$\sin \alpha = 1$,which implies $\alpha = \frac{\pi}{2}$. Checking with the coefficient: $-1 = 1 - 2$,which is $-1 = -1$. Thus,$\alpha = \frac{\pi}{2}$.

If $\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{\sin \alpha}{\cos x} - \frac{2}{1 + \cos x}$,then $\alpha = $

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