We will use the following Rules :
#(R1) : sintheta=cos(pi/2-theta)#
#(R2) : cos^-1(costheta)=theta, theta in [o,pi]#
#(R3) : sin^-1(sintheta)=theta, theta in [-pi/2,pi/2]#
Let us note that, by
#(R1), sin(pi/12)=cos(pi/2-pi/12)=cos(5pi/12)#
So, #cos^-1(sin(pi/12))=cos^-1(cos(5pi/12))#, where,
#5pi/12 in [0,pi]#, so, using #(R2)#, we get,
#cos^-1(cos(5pi/12))=5pi/12#
Again, as #5pi/12 in [-pi/2,pi/2]#, by #(R3)#, we have,
#sin^-1(sin(5pi/12))=5pi/12#
Finally, #sin^-1(sin(cos^-1(sin(pi/12))))=5pi/12#.
Hope, this will be of Help! Enjoy Maths.!