5) \( \frac{\cos (x)-1}{1+\cos (x)}-\frac{1+\cos (x)}{\cos (x)-1}=4 \cot (x) \csc (x) \)
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Let's start solving the given equation: \[ \frac{\cos(x) - 1}{1 + \cos(x)} - \frac{1 + \cos(x)}{\cos(x) - 1} = 4 \cot(x) \csc(x) \] First, we need a common denominator for the left-hand side. The common denominator is \((1 + \cos(x))(\cos(x) - 1)\). Thus, we rewrite each fraction: \[ \frac{(\cos(x) - 1)^2 - (1 + \cos(x))^2}{(1 + \cos(x))(\cos(x) - 1)} = 4 \cot(x) \csc(x) \] Next, we'll simplify the numerator: \[ (\cos(x) - 1)^2 - (1 + \cos(x))^2 = (\cos^2(x) - 2\cos(x) + 1) - (1 + 2\cos(x) + \cos^2(x)) \] \[ = \cos^2(x) - 2\cos(x) + 1 - 1 - 2\cos(x) - \cos^2(x) = -4\cos(x) \] Now, the left side becomes: \[ \frac{-4\cos(x)}{(1 + \cos(x))(\cos(x) - 1)} = 4 \cot(x) \csc(x) \] Thus, we need to solve: \[ \frac{-4\cos(x)}{(1 + \cos(x))(cos(x) - 1)} = 4 \cot(x) \csc(x) \] Next, let's substitute \(\cot(x)\) and \(\csc(x)\): \[ \cot(x) = \frac{\cos(x)}{\sin(x)}, \quad \csc(x) = \frac{1}{\sin(x)} \implies \cot(x) \csc(x) = \frac{\cos(x)}{\sin^2(x)} \] So, we can rewrite the equation: \[ \frac{-4\cos(x)}{(1 + \cos(x))(cos(x) - 1)} = \frac{4\cos(x)}{\sin^2(x)} \] We simplify this to: \[ - \frac{1}{(1 + \cos(x))(cos(x) - 1)} = \frac{1}{\sin^2(x)} \] Taking the reciprocal gives us: \[ (1 + \cos(x))(1 - \cos(x)) = - \sin^2(x) \] Notice that \(1 - \cos^2(x) = \sin^2(x)\), leading us to: \[ 1 - \cos^2(x) = - \sin^2(x) \] which indicates \(1 = 0\), an impossible scenario. Thus, the original equation does not hold true except perhaps under specific conditions or for specific values of \(x\). Solving it further or identifying values may provide additional insights or perpendicular conjectures, but generally, this algebra ends up being contradictory, signaling either an error in the original presupposition or an elementary constraint upon the variables involved. In conclusion, the equation doesn't hold true in general and should be approached with caution.
