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

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

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

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

$=x^2+1-x^2=1$