Question

How do you verify $\tan \left(\pi -x\right)=-\tan x$?

Answer 1

It is known that $\mathbf{tan}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{=}\frac{\mathbf{tan}\mathbf{}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{-}\mathbf{tan}\mathbf{}\mathbf{\left(}\mathbf{B}\mathbf{\right)}}{\mathbf{1}\mathbf{+}\mathbf{tan}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{\times }\mathbf{tan}\mathbf{\left(}\mathbf{B}\mathbf{\right)}}$.

Thus,

$\tan \left(\pi -x\right)=\frac{\tan \left(\pi \right)-\tan \left(x\right)}{1+\left[\tan \left(\pi \right)\times \tan \left(x\right)\right]}$.

$$\begin{gathered} \Rightarrow \tan \left(\pi -x\right)=\frac{\tan \left(\pi -0\right)-\tan \left(x\right)}{1+\left[\tan \left(\pi -0\right)\times \tan \left(x\right)\right]} \\ \Rightarrow \tan \left(\pi -x\right)=\frac{\tan \left(0\right)-\tan \left(x\right)}{1+\left[\tan \left(0\right)\times \tan \left(x\right)\right]} \\ \Rightarrow \tan \left(\pi -x\right)=\frac{0-\tan \left(x\right)}{1+\left[0\times \tan x\right]} \\ \Rightarrow \tan \left(\pi -x\right)=\frac{-\tan \left(x\right)}{1} \\ \Rightarrow \tan \left(\pi -x\right)=-\tan \left(x\right) \end{gathered}$$

Therefore, it is verified that $\tan \left(\pi -x\right)=-\tan \left(x\right)$.