a couple of cheap answers to this one will be
[tex]\bf \cfrac{sec(x+1)}{tan(x)}=\cfrac{tan(x)}{sec(x-1)} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{~~\frac{1}{cos(x+1)}~~}{\frac{sin(x)}{cos(x)}}\implies \cfrac{1}{cos(x+1)}\cdot \cfrac{cos(x)}{sin(x)}\implies \cfrac{cos(x)}{sin(x)[cos(x+1)]}\ne \cfrac{tan(x)}{sec(x-1)}[/tex]
another way to check that will be, doing both sides, left and right.
[tex]\bf \cfrac{sec(x+1)}{tan(x)}=\cfrac{tan(x)}{sec(x-1)}\implies sec(x+1)sec(x-1)=tan^2(x) \\\\\\ \begin{array}{llll} \cfrac{1}{cos(x+1)cos(x-1)}&=\cfrac{sin^2(x)}{cos^2(x)} \\\\\\ \cfrac{1}{[cos(x)cos(1)-sin(x)sin(1)]~~[cos(x)cos(1)+sin(x)sin(1)]} \\\\\\ \cfrac{1}{[cos(x)cos(1)]^2-[sin(x)sin(1)]^2} \\\\\\ \cfrac{1}{cos^2(x)cos^2(1)-sin^2(x)sin^2(1)}&\ne \cfrac{sin^2(x)}{cos^2(x)} \end{array}[/tex]
you can also try graphing both, the left and right side independently, and notice that there's no match.
