\begin{align*} \ln |\csc x-\cot x| &=-\ln|\csc x+\cot x|&&\\ &=\ln \Big|\frac{1}{\sin x}-\frac{\cos x}{\sin x}\Big| &&\text{Recipricol functions}\\ &=\ln \Big|\frac{1-\cos x}{\sin x}\Big| &&\text{Use common denominator}\\ &=\ln |1-\cos x|-\ln |\sin x| &&\text{Property of logarithms}\\ &=-(\ln |\sin x|-\ln |1-\cos x|) &&\text{Factor a -1}\\ &=-\ln \Big|\frac{\sin x}{1-\cos x}\Big| &&\text{Property of logarithms}\\ &=-\ln \Big|\frac{\sin x(1+\cos x)}{1-\cos^2 x}\Big| &&\text{Multiply by }1-\cos x\\ &=-\ln \Big|\frac{\sin x(1+ \cos x)}{\sin^2 x}\Big| &&\text{Pythagorean Identity}\\ &=-\ln \Big|\frac{1+ \cos x}{\sin x}\Big| &&\text{Simplify by } \sin x\\ &=-\ln \Big|\frac{1}{\sin x}+\frac{\cos x}{\sin x}\Big| &&\text{Split in two fractions}\\ &=-\ln |\csc x +\cot x| \end{align*}

\begin{align*} \frac{\cot x+\cot y}{1-\cot x \cot y} &=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1-\frac{\cos x}{\sin x}\cdot \frac{\cos y}{\sin y}} &&\\ &=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1-\frac{\cos x}{\sin x}\cdot \frac{\cos y}{\sin y}} \cdot \frac{\sin x\sin y}{\sin x \sin y}\\ &=\frac{\frac{\cos x}{\sin x}\cdot \frac{\sin x\sin y}{1}+\frac{\cos y}{\sin y}\cdot \frac{\sin x\sin y}{1}}{\sin x \sin y-\frac{\cos x}{\sin x}\cdot \frac{\cos y}{\sin y}\cdot \frac{\sin x\sin y}{1}} \\ &=\frac{\cos x \sin y + \sin x \cos y}{\sin x\sin y - \cos x \cos y} \end{align*}

\begin{align*} \frac{\sin x}{1-\cot x}-\frac{\cos x}{\tan x-1} &=\sin x+\cos x &&\\ &=\frac{\sin x}{1-\frac{\cos x}{\sin x}}-\frac{\cos x}{\frac{\sin x}{\cos x}-1} &&\text{Apply fundamental identities}\\ &=\frac{\sin x}{\frac{\sin x-\cos x}{\sin x}}-\frac{\cos x}{\frac{\sin x-\cos x}{\cos x}} &&\text{Combine common denominator}\\ &= \frac{\sin^2 x}{\sin x -\cos x}-\frac{\cos^2 x}{\sin x -\cos x}&&\text{Divide Fractions}\\ &=\frac{\sin^2 x-\cos^2 x}{\sin x-\cos x}&&\text{Combine common denominator}\\ &=\frac{(\sin x+\cos x)(\sin x- \cos x)}{\sin x -\cos x} &&\text{Factor out numerator}\\ &= \sin x+\cos x &&\text{Simplify} \end{align*}

\begin{align*} \frac{\sin 4x+\sin 2x}{\cos 4x +\cos 2x} &= \frac{2 \sin \frac{1}{2}(4x+2x)\cos \frac{1}{2}(4x-2x)}{2\cos \frac{1}{2}(4x+2x)\cos \frac{1}{2}(4x-2x)} &&\text{Sum and Difference of Sines and Cosines}\\ &=\frac{\sin 3x}{\cos 3x} &&\text{Simplify}\\ &=\tan3x &&\text{Fundamental Identity} \end{align*}

Solve for $\cos \theta$ \begin{align*} 2\cos^2 \theta+\cos \theta-1 & =0 && \text{Original Eqaution.}\\ 2y^2+y-1 &=0 &&\text{Replace }y=\cos \theta.\\ (2y-1)(y+1) &=0 &&\text{Factor the quadratic expression.}\\ 2y-1=0 \text{ or } y+1&=0 &&\text{Set each factor equal to 0.}\\ y=\frac{1}{2} \text{ or } y&=-1 &&\text{Solve each resulting equation for } y.\\ \cos \theta=\frac{1}{2} \text{ or } \cos \theta&=-1 &&\text{Substitute back to } \theta: y=\cos \theta. \end{align*} Find the values of $\theta$ on $0 \leq \theta < 2 \pi$ that satisfy the equation $\cos \theta=\frac{1}{2}$. The cosine function is positive in quadrants I and IV and angles which yields $\cos \theta=\frac{1}{2}$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.
Find the values of $\theta$ on $0 \leq \theta < 2 \pi$ that satisfy the equation $\cos \theta=-1$. The only angle in $[0, 2\pi)$ which satisfy $\cos \theta=-1$ is $\pi$
The solutions to $2\cos^2 \theta-\cos \theta-1=0$ are: $$\theta=\frac{\pi}{3}, \frac{5\pi}{3} \text{ or }\theta=\pi$$

Note that $\sin x-1$ must be greater than 0 since domain of logarithm functions is $x>0$. This yield in $\sin x>1$. \begin{align*} & \ln(2\sin x)+\ln(\sin x -1)=\ln(\sin x-1) && \\ &\ln[2\sin x(\sin x -1)]=\ln (\sin x-1) &&\text{Property of the logarithms}\\ &\ln (2\sin^2 x-2\sin x)=\ln (\sin x-1) &&\text{Distributive property}\\ &2\sin^2 x-2\sin x=\sin x-1 &&\text{Property of the logarithms}\\ &2\sin^2 x-3\sin x+1=0 &&\text{Simplify like terms}\\ &(2\sin x-1)(\sin x-1)=0 &&\text{Factoring}\\ &x=\frac{\pi}{6}, x=\frac{5\pi}{6} \text{ or } x=\frac{\pi}{2} &&\text{Solving each factor}\\ \end{align*} However all these values are extraneous since they are causing a domain error.