Question #67a2f

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Answer 1 · 2017-05-13T11:49:40

The answer is $=u/sqrt(4+u^2)$

We need

$sin^2x+cos^2x=1$

$1/cos^2x=1+tan^2x$

Let,

$y=tan^-1(u/2)$

$tany=u/2$

$1/cos^2y=1+u^2/4=(4+u^2)/4$

$cos^2y=4/(4+u^2)$

$sin^2y=1-cos^2y=1-4/(4+u^2)$

$=((4+u^2)-4)/(4+u^2)=u^2/(4+u^2)$

$siny=u/sqrt(4+u^2)$

Here, we are looking at

$sin(tan^-1(u/2))=siny=u/sqrt(4+u^2)$