5) $$ \frac{\cos (x)-1}{1+\cos (x)}-\frac{1+\cos (x)}{\cos (x)-1}=4 \cot (x) \csc (x) $$

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Solve the equation $$ \frac{\cos (x)-1}{1+\cos (x)}-\frac{1+\cos (x)}{\cos (x)-1}=4 \cot (x) \csc (x) $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{\cos\left(xight)-1}{1+\cos\left(xight)}-\frac{1+\cos\left(xight)}{\cos\left(xight)-1}=4\cot\left(xight)\csc\left(xight)$$
- step1: Find the domain:
  $$\frac{\cos\left(xight)-1}{1+\cos\left(xight)}-\frac{1+\cos\left(xight)}{\cos\left(xight)-1}=4\cot\left(xight)\csc\left(xight),x
eq k\pi ,k \in \mathbb{Z}$$
- step2: Rewrite the expression:
  $$\frac{\cos\left(xight)-1}{1+\cos\left(xight)}-\frac{1+\cos\left(xight)}{\cos\left(xight)-1}=4	imes \frac{\cos\left(xight)}{\sin\left(xight)}	imes \frac{1}{\sin\left(xight)}$$
- step3: Simplify:
  $$\frac{\cos\left(xight)-1}{1+\cos\left(xight)}-\frac{1+\cos\left(xight)}{\cos\left(xight)-1}=\frac{4\cos\left(xight)}{\sin^{2}\left(xight)}$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{\cos\left(xight)-1}{1+\cos\left(xight)}-\frac{1+\cos\left(xight)}{\cos\left(xight)-1}ight)\left(1+\cos\left(xight)ight)\left(\cos\left(xight)-1ight)\sin^{2}\left(xight)=\frac{4\cos\left(xight)}{\sin^{2}\left(xight)}	imes \left(1+\cos\left(xight)ight)\left(\cos\left(xight)-1ight)\sin^{2}\left(xight)$$
- step5: Simplify the equation:
  $$\left(\cos\left(xight)-1ight)\left(\cos\left(xight)-1ight)\sin^{2}\left(xight)+\left(-1-\cos\left(xight)ight)\left(1+\cos\left(xight)ight)\sin^{2}\left(xight)=4\cos\left(xight)\left(1+\cos\left(xight)ight)\left(\cos\left(xight)-1ight)$$
- step6: Calculate:
  $$\left(\cos^{2}\left(xight)-2\cos\left(xight)+1ight)\sin^{2}\left(xight)+\left(-1-2\cos\left(xight)-\cos^{2}\left(xight)ight)\sin^{2}\left(xight)=4\cos\left(xight)\left(1+\cos\left(xight)ight)\left(\cos\left(xight)-1ight)$$
- step7: Rewrite the expression:
  $$-4\cos\left(xight)+4\cos^{3}\left(xight)=-4\cos\left(xight)+4\cos^{3}\left(xight)$$
- step8: Cancel equal terms:
  $$0=0$$
- step9: The statement is true:
  $$x \in \mathbb{R}$$
- step10: Check if the solution is in the defined range:
  $$x \in \mathbb{R},x
eq k\pi ,k \in \mathbb{Z}$$
- step11: Find the intersection:
  $$x
eq k\pi ,k \in \mathbb{Z}$$
The solution to the equation $$ \frac{\cos (x)-1}{1+\cos (x)}-\frac{1+\cos (x)}{\cos (x)-1}=4 \cot (x) \csc (x) $$ is $$ x 
eq k\pi, k \in \mathbb{Z} $$.
The solution to the equation is all real numbers except integer multiples of π.

5) \( \frac{\cos (x)-1}{1+\cos (x)}-\frac{1+\cos (x)}{\cos (x)-1}=4 \cot (x) \csc (x) \)

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