Let sin^-1x=theta", where, "|x|le1.
Then, by the Defn. of sin^-1 fun.,
sintheta=x, &, theta in [-pi/2,pi/2].
theta in [-pi/2,pi/2] rArr -pi/2le thetalepi/2.
Multiplying the inequality by -1, hence, reversing its order,
pi/2ge-thetage-pi/2. Now, adding, pi/2,
pigepi/2-thetage0, i.e., (pi/2-theta) in [0,pi].
Further, cos(pi/2-theta)=sintheta=x.
Thus, cos(pi/2-theta)=x, where, (pi/2-theta) in [0,pi].
This, together with the Defn. of cos^-1 fun., allow us to say that,
(pi/2-theta)=cos^-1x.
Replacing theta by sin^-1x, we finally arrive at,
pi/2-sin^-1x=cos^-1x, or, sin^-1x+cos^-1x=pi/2.
Enjoy maths.!
