We can offer two approaches to this problem - using trigonometry of right triangles (applicable for positive #x#) and purely trigonometric (applicable for all #x#, but we will use it for #x<=0#.

Let's analyze this problem from the position of trigonometry of the right triangle. For this, let's assume that #0<x<=1#.

Then #/_A=arcsin x# is an angle from #0# to #pi/2#, *sine* of which is #x#.

Assume, this angle #/_A# is an acute angle of a right triangle #Delta ABC#. So, #sin(/_A)# is a ratio of an opposite cathetus #a# to hypotenuse #c#:

#x=sin(/_A)=a/c#

Consider another acute angle of this triangle - #/_B#.

Expression #a/c# represents its *cosine*.

Therefore,

#/_B = arccos(a/c)=arccos x#

Now we see that #/_A=arcsin x# and #/_B=arccose x# are two acute angles in a right triangle. Their sum is always #pi/2#.

Therefore,

#sin(arcsin x+arccos x)=sin(pi/2)=1#

Case with non-positive #x# we will consider differently.

If #x <= 0#, #arcsin x# is between #-pi/2# and #0#.

If #x <= 0#, #arccos x# is between #pi/2# and #pi#.

Using formula for #sin# of a sum of two angles,

#sin(arcsin x+arccos x) =#

#= sin(arcsin x)*cos(arccosx) + cos(arcsin x)*sin(arccos y)#

By definition of inverse trigonometric functions #arcsin# and #arccos#, we write the following:

#sin(arcsin x)=x#

#cos(arccos x)=x#

Using trigonometric identity #sin^2(phi)+cos^2(phi)=1#, we can find the other components:

#cos^2(arcsin x) = 1-sin^2(arcsinx) = 1-x^2#

#sin^2(arccos x) = 1-cos^2(arccos x) = 1-x^2#

Since #arcsin x# is between #-pi/2# and #0#, its *cosin* is positive:

#cos^2(arcsin x) = 1-x^2#

#=>cos(arcsin x) sqrt(1-x^2)#

Since #arccos x# is between #pi/2# and #pi#, its *sine* is positive:

#sin^2(arccos x) = 1-x^2#

#=>sin(arccos x) sqrt(1-x^2)#

Now we can calculate the value of the original expression:

#sin(arcsin x+arccos x) =#

#= sin(arcsin x)*cos(arccosx) + cos(arcsin x)*sin(arccos y) =#

#= x*x+sqrt(1-x^2)*sqrt(1-x^2) =#

#= x^2 +1 -x^2 = 1#

(as in the case we did geometrically).