## Precalculus (6th Edition) Blitzer

b) In order to verify the equations, use the trigonometric identities. $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$ Now, apply the above identities in the following equation: \begin{align} & \frac{\tan x-1}{1-\cot x}=\frac{\frac{\sin x}{\cos x}-1}{1-\frac{\cos x}{\sin x}} \\ & =\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{\sin x-\cos x}{\sin x}} \\ & =\frac{\left( \sin x-\cos x \right)\sin x}{\left( \sin x-\cos x \right)\cos x} \\ & =\frac{\sin x}{\cos x} \end{align} By using $\tan x=\frac{\sin x}{\cos x}$, we get: \begin{align} & \frac{\tan x-1}{1-\cot x}=\frac{\sin x}{\cos x} \\ & =\tan x \end{align} Hence, the equation $\frac{\tan x-1}{1-\cot x}$ is equal to $\tan x$.