## Precalculus (6th Edition) Blitzer

The two conditions $\tan \theta <0$ and $\cos \theta >0$ meet in quadrant IV.
Assume $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ is the distance from $\left( 0,0 \right)$ to $\left( x,y \right)$ Here, $r$ is positive in all quadrants. Also, $x$ and $y$ can be positive or negative. The cosine function is defined as $\cos \theta =\frac{x}{r}$. The value $x$ is positive in quadrant I and IV. The value of $\cos \theta$ is positive in quadrant I and IV. The definition of the tangent function is $\tan \theta =\frac{y}{x}$. In quadrant I and IV either the $x$ or $y$ - value is negative. Therefore, $\tan \theta$ is negative in quadrants II and IV. And the two conditions $\tan \theta <0$ and $\cos \theta >0$ meet in quadrant IV. Thus, $\theta$ must lie in quadrant IV.