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.!
