#r = 10csc(theta + (7pi)/4)#

Multiply by sine.

#r sin(theta + (7pi)/4) = 10#

Now use the sine addition formula:

#sin(theta + (7pi)/4) = sin(theta)cos((7pi)/4) + cos(theta)sin((7pi)/4)#

Since #(7pi)/4# is in quadrant IV, sine is negative and cosine is positive, with

#sin((7pi)/4) = -sqrt(2)/2#

and

#cos((7pi)/4) = sqrt(2)/2#

Therefore we have...

#r sin(theta + (7pi)/4) = 10#

#rsin(theta)cos((7pi)/4) + rcos(theta)sin((7pi)/4) = 10#

#-rsin(theta)(sqrt(2)/2) + rcos(theta)(sqrt(2)/2)= 10#

Multiply both sides by #sqrt(2)#. This has the same effect as dividing by #sqrt(2)/2#.

#-rsin(theta) + rcos(theta)= 10sqrt(2)#

Now use the standard polar substitutions:

#x = rcos(theta) and y = rsin(theta)#.

We have...

#-y + x = 10sqrt(2)#

Solve for y, or put it in whatever form you desire. This is a line.