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

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(C) Given the equation: $\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{\sin \alpha}{\cos x} - \frac{2}{1 + \cos x}$.We can simplify the left side usin

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$\frac{\pi}{2}$

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(C) Given the equation: $\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{\sin \alpha}{\cos x} - \frac{2}{1 + \cos x}$.We can simplify the left side using partial fractions or by algebraic manipulation.Let $\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{A}{\cos x} + \frac{B}{1 + \cos x}$.$1 - \cos x = A(1 + \cos x) + B(\cos x)$.If $\cos x = 0$,then $1 = A(1) \implies A = 1$.If $\cos x = -1$,then $1 - (-1) = B(-1) \implies 2 = -B \implies B = -2$.So,$\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{1}{\cos x} - \frac{2}{1 + \cos x}$.Comparing this with the given equation $\frac{\sin \alpha}{\cos x} - \frac{2}{1 + \cos x}$,we get $\sin \alpha = 1$.Therefore,$\alpha = \frac{\pi}{2}$.

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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