Question #bf76e

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Answer 1 · 2016-08-01T08:18:13

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

#=>cos^-1x=pi/2-A=pi/2-sin^-1x#

#sin^-1x+cos^-1x=pi/2#

So the given expression

#sin(sin^-1x+cos^-1x)=sin(pi/2)=1#

Alternative way

#sin(sin^-1x+cos^-1x)#

#=(sinsin^-1x)(coscos^-1x)+(cossin^-1x)(sincos^-1x)#

#=x*x+(coscos^-1sqrt(1-x^2))(sinsin^-1sqrt(1-x^2))#

#=x^2+1-x^2=1#