If the apparent dip circle is at an angle $\alpha$ with true dip circle then,

$\tan \left(\varphi \text{'}\right)>\tan \left(\varphi \right) \Rightarrow \varphi \text{'}>\varphi$

Therefore, apparent dip angle is greater than true dip angle.

Hence, true dip $\left(\varphi \right)$ is less than apparent dip $\left({\varphi }^{\text{'}}\right)$