Question #bf76e

1 Answer
Aug 1, 2016

Let sin^-1x=Asin1x=A
=>x=sinA=cos(pi/2-A)x=sinA=cos(π2A)

=>cos^-1x=pi/2-A=pi/2-sin^-1xcos1x=π2A=π2sin1x

sin^-1x+cos^-1x=pi/2sin1x+cos1x=π2

So the given expression

sin(sin^-1x+cos^-1x)=sin(pi/2)=1sin(sin1x+cos1x)=sin(π2)=1

Alternative way

sin(sin^-1x+cos^-1x)sin(sin1x+cos1x)

=(sinsin^-1x)(coscos^-1x)+(cossin^-1x)(sincos^-1x)=(sinsin1x)(coscos1x)+(cossin1x)(sincos1x)

=x*x+(coscos^-1sqrt(1-x^2))(sinsin^-1sqrt(1-x^2))=xx+(coscos11x2)(sinsin11x2)

=x^2+1-x^2=1=x2+1x2=1