#### Answer

The two conditions $\tan \theta <0$ and $\cos \theta >0$ meet in quadrant IV.

#### Work Step by Step

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.