Question #2dd5b

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Answer 1 · 2017-02-15T18:35:19

Given $tanu=5/12$
Both $u and v$ are in III quadrant
So
$tanu->+ve$

$tanv->+ve$

$sinu->-ve$

$sinv->-ve$

$cosu->-ve$

$cosv->-ve$

$sinu=1/cscu=-1/sqrt(1+cot^2u)=-1/sqrt(1+12^2/5^2)=-5/13$

$cosu=sinu/tanu=-(5/13)/(5/12)=-12/13$

Given $sinv=-3/5$

$cosv$
$=-sqrt(1-sin^2v)=-sqrt(1-9/25)=-4/5$

$tanv=sinv/cosv=3/4$

$sin(u+v)$

$=sinucosv+cosusinv$

$=(-5/13)(-4/5)+(-12/13)(-3/5)$

$=20/65+36/65=56/65$

$tan(u+v)=(tanu+tanv)/(1-tanutanv)$

$=(5/12+3/4)/(1-5/12*3/4)=(14/12)/(11/16)=7/6*16/11=56/33$

$cos(u+v)$

$=cosucosv-sinusinv$

$=(-12/13)(-4/5)-(-5/13)(-3/5)$

$=48/65-15/65=33/65$