Let sin^-1x=Asin−1x=A
=>x=sinA=cos(pi/2-A)⇒x=sinA=cos(π2−A)
=>cos^-1x=pi/2-A=pi/2-sin^-1x⇒cos−1x=π2−A=π2−sin−1x
sin^-1x+cos^-1x=pi/2sin−1x+cos−1x=π2
So the given expression
sin(sin^-1x+cos^-1x)=sin(pi/2)=1sin(sin−1x+cos−1x)=sin(π2)=1
Alternative way
sin(sin^-1x+cos^-1x)sin(sin−1x+cos−1x)
=(sinsin^-1x)(coscos^-1x)+(cossin^-1x)(sincos^-1x)=(sinsin−1x)(coscos−1x)+(cossin−1x)(sincos−1x)
=x*x+(coscos^-1sqrt(1-x^2))(sinsin^-1sqrt(1-x^2))=x⋅x+(coscos−1√1−x2)(sinsin−1√1−x2)
=x^2+1-x^2=1=x2+1−x2=1