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)#