## Precalculus (6th Edition) Blitzer

$-1$
Since, $\sec^2 \theta =1+\tan^2 \theta$ Here, $\theta = 1.4^{\circ}$ We have: $\tan^2 1.4^{\circ}-\sec^2 1.4^{\circ}$ or, $= \tan^2 1.4^{\circ}-(1+\tan^2 1.4^{\circ} )$ or, $=-1$