Question #02bba

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Answer 1 · 2016-11-02T05:00:11

Given $0< u< pi/2$ and $tanu=5/3$

we get $u=tan^-1(5/3)~~59^@$

So $2u=118^@->"In 2nd quadrant"$

Hence $sin2u->"+ve"$

But $tan2u and cos 2u ->"-ve"$

Now

$sin2u=(2tanu)/(1+tan^2u)$

$=>sin2u=(2xx5/3)/(1+(5/3)^2)=10/3xx9/34=15/17$

$cos2u=-(1-tan^2u)/(1+tan^2u)$

$=>cos2u=(1-(5/3)^2)/(1+(5/3)^2$

$=>cos2u=-16/9xx9/34-8/17$

$tan2u=(2tanu)/(1-tan^2u)$

$=>tan2u=(2xx5/3)/(1-(5/3)^2)=-10/3xx9/16=-15/8$