Solve for t algebraically: inverse cos(t) = inverse sin(t).
Where do I start?
since sin(x) = cos(π/2-x)
arccos(t) + arcsin)t) = π/2
In general, draw the triangle. In this case, take cosine of both sides.
cos(arccos(t)) = t
In your triangle, the adjacent side will be t, and the hypotenuse will be 1. The other leg is thus √(1-t^2). So,
sin(arccos(t)) = √(1-t^2)
Now we have
t = √(1-t^2)
t^2 = 1-t^2
2t^2 = 1
t = 1/√2
So, arccos(1/√2) = arcsin(1/√2) = π/4
So, their sum is π/4 + π/4 = π/2
You can sometimes save yourself a lot of work by remembering a few fundamental relationships.