## Precalculus (6th Edition) Blitzer

Let us consider the left side of the given expression: $\tan \left( 2\pi -x \right)$ By using the identity of trigonometry, $\tan \,\left( \alpha -\beta \right)=\frac{\tan \,\alpha -\tan \,\beta }{1+\tan \,\alpha \tan \,\beta }$, the above expression can be further simplified as: \begin{align} & \tan \left( 2\pi -x \right)=\frac{\tan \,2\pi -\tan \,x}{1+\tan \,2\pi \tan \,x} \\ & =\frac{0-\tan \,x}{1+0\times \tan \,x} \\ & =\frac{-\tan \,x}{1} \\ & =-\tan \,x \end{align} Hence, the the left side of the provided expression is equal to the right side, which is $\tan \left( 2\pi -x \right)=-\tan \,x$.