How do you simplify #2 +tan^2(x)/sec^2(x)-1#?

#(2 + tan ^2x) / (sec^2 x - 1) = (1 + (1 + tan^2 x)) / (sec^x - 1)#

# = (1 + sec^2 x) / (sec ^2 x - 1)#

#= (cos^2 x + 1) / cancel(cos^2 x) / (1-cos^2x)/cancel(cos^2x)#

#= (1 + cos^2x) / sin^2x =color(red)( csc^2x + cot^2x)#

Notes:
[1] I have replaced the #x#'s in the original expression with #theta#'s because I wanted to use #x# in the context of the coordinate plane.
[2] I have use the form of the expression indicated in the comment (below).
[3] Because we have been given an expression (and not an equation) this can be simplified but it can not be solved.

To more easily see the relations involved consider the standard trig ratio triangle:
#color(white)("XXX")tan(theta)=(Deltax)/(Deltay)#

#color(white)("XXX")cos(theta)=(Deltay)/(Deltax)color(white)("xx")rarrcolor(white)("xx")sec(theta)=(Deltar)/(Deltay)#

#color(white)("XXX")sin(theta)=(Deltay)/(Deltar)#

Working through the given expression a small piece at a time:
#(tan(theta))/(sec(theta))=((Deltax)/(Deltay))/((Deltar)/(Deltay))=(Deltax)/(Deltar)=sin(theta)#

#(tan^2(theta))/(sec^2(theta))=((tan(theta))/(sec(theta)))^2=sin^2(theta)#

#(tan^2(theta))/(sec^2(theta))-1=sin^2(theta)-1#

#color(white)("XXXXXXX")=-cos(theta)color(white)("xxxxxxxx")# since #cos^2(theta)+sin^2(theta)=1#

#2+(tan^2(theta))/(sin^2(theta))-1 = 2-cos^2(theta)#

#color(white)("xxxxxxxxxxxxx")=1+(1-cos^2(theta))#

#color(white)("xxxxxxxxxxxxx")=1+sin^2(theta)#