Chapter 7.5, Problem 53AYU

establish each identify.

$\tan \left(\pi -\theta \right)\text{ = }-\text{tan}\theta$

To determine

To establish: The identity $\text{tan}\left(\pi -\theta \right)=-\tan \theta$.

Answer to Problem 53AYU

Solution:

The identity $\text{tan}\left(\pi -\theta \right)=-\tan \theta$ is established.

Explanation of Solution

Given:

$\text{tan}\left(\pi -\theta \right)=-\tan \theta$

Formula used:

$\text{tan}\left(\pi -\theta \right)=\frac{\sin \left(\pi -\theta \right)}{\cos \left(\pi -\theta \right)}$

Calculation:

Left hand side

$\text{tan}\left(\pi -\theta \right)$

$\alpha =\pi ,\beta =\theta$

By difference formula of $\text{sine}$ and $\text{cosine}$

$\text{tan}\left(\pi -\theta \right)=\frac{\sin \pi \cos \theta -\cos \pi \sin \theta }{\cos \pi \cos \theta +\sin \pi \sin \theta }$

$\text{tan}\left(\pi -\theta \right)=\frac{0.\cos \theta -\left(-1\right)\sin \theta }{\left(-1\right)\cos \theta +0.\sin \theta }$

$\text{tan}\left(\pi -\theta \right)=\frac{\sin \theta }{-\cos \theta }$

$\text{tan}\left(\pi -\theta \right)=-\tan \theta$

$\text{LHS = RHS}$