## Precalculus (6th Edition) Blitzer

$1$
Solve. $\csc 37^{\circ} \sec 53^{\circ}-\tan 53^{\circ} \cot 37^{\circ}$ We have $= \sec (90^{\circ}-37^{\circ}) \sec 53^{\circ}-\tan 53^{\circ} \tan 53^{\circ}$ or, $=\sec^2 53^{\circ}-\tan^2 53^{\circ}$ Since, $1+\tan^2 x=\sec^2 x$ So, the answer is : $=1$