## Precalculus (6th Edition) Blitzer

The expression in terms of the function $\csc x$ is $2{{\csc }^{2}}x-1$.
By multiplying the part of expression $\frac{1}{1-\cos x}$ by $\frac{1+\cos x}{1+\cos x}$ and the other part of expression $\frac{\cos x}{1+\cos x}$ by $\frac{1-\cos x}{1-\cos x}$: \begin{align} & \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=\frac{1}{1-\cos x}\times \frac{1+\cos x}{1+\cos x}-\frac{\cos x}{1+\cos x}\times \frac{1-\cos x}{1-\cos x} \\ & =\frac{1+\cos x}{1-{{\cos }^{2}}x}-\frac{\cos x-{{\cos }^{2}}x}{1-{{\cos }^{2}}x} \\ & =\frac{1+\cos x-\cos x+{{\cos }^{2}}x}{1-{{\cos }^{2}}x} \\ & =\frac{1+{{\cos }^{2}}x}{1-{{\cos }^{2}}x} \end{align} Apply the Pythagorean identity of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. And the expression $1-{{\cos }^{2}}x$ can be written as $si{{n}^{2}}x$. Then, $\frac{1+{{\cos }^{2}}x}{1-{{\cos }^{2}}x}=\frac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}$ Now, the above expression is further simplified as, $\frac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}=\frac{1}{{{\sin }^{2}}x}+\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$ By using the reciprocal identity of trigonometry $\csc x=\frac{1}{\sin x}$ , and the quotient identity of trigonometry ${{\cot }^{2}}x=\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$. And the above expression can be further simplified as, $\frac{1}{{{\sin }^{2}}x}+\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x}={{\csc }^{2}}x+{{\cot }^{2}}x$ By using the Pythagorean identity of trigonometry ${{\csc }^{2}}x=1+{{\cot }^{2}}x$. Then, The above expression is simplified as, \begin{align} & {{\csc }^{2}}x+{{\cot }^{2}}x={{\csc }^{2}}x+{{\csc }^{2}}x-1 \\ & =2{{\csc }^{2}}x-1 \end{align} Thus, the provided expression is written in the terms of the function $\csc x$ as $2{{\csc }^{2}}x-1$ Hence, the $\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}$ expression in terms of the function $\csc x$ is $2{{\csc }^{2}}x-1$.