Question 405450
<pre><font face = "batangche" color = "indigo" size = 4><b>

{{{drawing(400,400,-.5,.5,-.5,.5, graph(400,400,-.5,.5,-.5,.5),

arc(0,0,.4,-.4,0,180), arc(-.023,0,.35,-.35,180,360),

arc(0,0,.3,-.3,0,135), line(0,0,.4cos(3pi/4),.4sin(3pi/4)),
locate(0,.25,"495°")

  )}}} 

The angle 495° is more than one complete revolution counter-clockwise
from the right side of the x-axis.  So we subtract 1 revolution or
360° from 495° getting 135°.  So we can forget the first revolution.
We now have just this:

{{{drawing(400,400,-.5,.5,-.5,.5, graph(400,400,-.5,.5,-.5,.5),

arc(0,0,.3,-.3,0,135), line(0,0,.4cos(3pi/4),.4sin(3pi/4)),

locate(0,.2,"135°")

  )}}}

Next we see that 135° is in the second quadrant. Let's indicate the 
reference angle in red, which is the smallest possible amout of 
rotation to get to the x-axis from the terminal side:

{{{drawing(400,400,-.5,.5,-.5,.5, graph(400,400,-.5,.5,-.5,.5),

arc(0,0,.3,-.3,0,135), line(0,0,.4cos(3pi/4),.4sin(3pi/4)),
red(arc(0,0,.28,-.28,135,180)),
locate(0,.2,"135°")

  )}}}

We subtract 135° from 180° to get 45°, which is the reference angle
indicated by the red arc.

{{{drawing(400,400,-.5,.5,-.5,.5, graph(400,400,-.5,.5,-.5,.5),

arc(0,0,.3,-.3,0,135), line(0,0,.4cos(3pi/4),.4sin(3pi/4)),
red(arc(0,0,.28,-.28,135,180), locate(-.19,.08,"45°")),
locate(0,.2,"135°")

  )}}}

Now we remember our 45°-45°-90° right triangle:

{{{drawing(200,200,0,1,0,1,triangle(0,0,1,1,1,0),
locate(.5,.1,1), locate(.9,.5,1), locate(.5,.5,sqrt(2)),
arc(0,0,.7,-.7,0,45), locate(.1,.1,"45°")

 )}}}

The cosecant is the hypotenuse over the opposite which
is {{{sqrt(2)/1}}} or just {{{sqrt(2)}}}.  We remember that
angles in the second quadrant have positive sines and cosecants,
so the answer is
               _
csc(495°) = +<font face = "symbol">Ö</font>2

Edwin</pre>