Answer
$$1$$
Work Step by Step
Recall the co-function identity:
$ \tan (\theta)=\cot(90^{\circ}-\theta)$ and $ \sec (\theta)=\csc (90^{\circ}-\theta)$
We know that
$\sec^2 \theta -\tan^2 \theta=1$
Therefore, $$\sec \ 35^{\circ} \csc 55^{\circ}-\tan 35^{\circ} \ \cot \ 55^{\circ} =\sec 35^{\circ} \ \sec (90^{\circ}-55^{\circ}) -\tan 35^{\circ} \tan (90^{\circ}-55^{\circ}) \\=\sec^2 \ 35^{\circ} -\tan^2 35^{\circ} \\=1$$