Question #2dd5b

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Question #2dd5b

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Answer 1 · 2017-02-15T18:35:19

Given $tan u = \frac{5}{12}$ $tanu=5/12$
Both $u and v$ $u and v$ are in III quadrant
So
$tan u \to + v e$ $tanu->+ve$

$tan v \to + v e$ $tanv->+ve$

$sin u \to - v e$ $sinu->-ve$

$sin v \to - v e$ $sinv->-ve$

$cos u \to - v e$ $cosu->-ve$

$cos v \to - v e$ $cosv->-ve$

$sin u = \frac{1}{csc u} = - \frac{1}{\sqrt{1 + {cot}^{2} u}} = - \frac{1}{\sqrt{1 + \frac{12^{2}}{5^{2}}}} = - \frac{5}{13}$ $sinu=1/cscu=-1/sqrt(1+cot^2u)=-1/sqrt(1+12^2/5^2)=-5/13$

$cos u = \frac{sin u}{tan u} = - \frac{\frac{5}{13}}{\frac{5}{12}} = - \frac{12}{13}$ $cosu=sinu/tanu=-(5/13)/(5/12)=-12/13$

Given $sin v = - \frac{3}{5}$ $sinv=-3/5$

$cos v$ $cosv$
$= - \sqrt{1 - {sin}^{2} v} = - \sqrt{1 - \frac{9}{25}} = - \frac{4}{5}$ $=-sqrt(1-sin^2v)=-sqrt(1-9/25)=-4/5$

$tan v = \frac{sin v}{cos v} = \frac{3}{4}$ $tanv=sinv/cosv=3/4$

$sin (u + v)$ $sin(u+v)$

$= sin u cos v + cos u sin v$ $=sinucosv+cosusinv$

$= (- \frac{5}{13}) (- \frac{4}{5}) + (- \frac{12}{13}) (- \frac{3}{5})$ $=(-5/13)(-4/5)+(-12/13)(-3/5)$

$= \frac{20}{65} + \frac{36}{65} = \frac{56}{65}$ $=20/65+36/65=56/65$

$tan (u + v) = \frac{tan u + tan v}{1 - tan u tan v}$ $tan(u+v)=(tanu+tanv)/(1-tanutanv)$

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

$cos (u + v)$ $cos(u+v)$

$= cos u cos v - sin u sin v$ $=cosucosv-sinusinv$

$= (- \frac{12}{13}) (- \frac{4}{5}) - (- \frac{5}{13}) (- \frac{3}{5})$ $=(-12/13)(-4/5)-(-5/13)(-3/5)$

$= \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$ $=48/65-15/65=33/65$

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