Question

Calcule a cotg ( arc sec (-3) + arc sen (0,1))

Answer 1

Calcule a cotg ( arc sec (-3) + arc sen (0,1))
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por partes :
arc sec (-3) = arc cos (-1/3) não é elementar. Na calculadora = 109,47º
arc sen (0,1) (tambem só com calculad.) = 5,74º
cotg (109,47 + 5,74) =
cotg 115,21º =
= 1/ tg 115,21º (só calculad.) = 1/(-2,124) = - 0,4708 (resp)

Answer 2

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$\sf cotg(arc~sec(-3)+arc~sen(0,1))\\\sf arc~sec(-3)=a\implies sec(a)=-3\\\sf arc~sen(0,1)=b\implies sen(b)=0,1\\\sf cotg(arc~sec(-3)+arc~sen(0,1))=cotg(a+b)\\\sf cotg(a+b)=\dfrac{cotg(a)\cdot cotg(b)-1}{cotg(a)\cdot cotg(b)}.$

$\underline{\rm c\acute alculos~auxiliares:}\\\sf tg^2(a)=sec^2(a)-1\\\sf tg^2(a)=(-3)^2-1\\\sf tg^2(a)=9 -1\\\sf tg^2(a)=8 \\\sf tg(a)=-\sqrt{8}=-2\sqrt{2}\\\boxed{\sf cotg(a)=\dfrac{1}{tg(a)}=\dfrac{1}{2\sqrt{2}}=-\dfrac{\sqrt{2}}{4}}\\\sf sen(b)=0,1=\dfrac{1}{10}\implies cosec(b)=10\\\sf cotg^2(b)=cosec^2(b)-1\\\sf cotg^2(b)=10^2-1\\\sf cotg^2(b)=100-1\\\sf cotg^2(b)=99$

$\boxed{\sf cotg(b)=\sqrt{99}=3\sqrt{11}}$

$\underline{\rm fazendo~as~substituic_{\!\!,}\tilde oes~ temos:}\\\sf cotg(a+b)=\dfrac{\frac{-\sqrt{2}}{4}+3\sqrt{11}-1}{\frac{\sqrt{2}}{4}\cdot3\sqrt{11}}\\\sf cotg(a+b)=\dfrac{\frac{-\sqrt{2}+12\sqrt{11}-4}{4}}{\frac{3\sqrt{22}}{4}}\\\sf cotg(a+b)=\dfrac{(-\sqrt{2}+12\sqrt{11}-4)}{\diagdown\!\!\!\!4}\cdot\dfrac{\diagdown\!\!\!\!4}{3\sqrt{22}}$

$\sf cotg(a+b)=\dfrac{(-\sqrt{2}+12\sqrt{11}-4)}{3\sqrt{22}}\cdot\dfrac{\sqrt{22}}{\sqrt{22}}\\\sf cotg(a+b)=\dfrac{-\sqrt{4\cdot11}+12\sqrt{2\cdot11^2}-4\sqrt{22}}{3\cdot22}\\\sf cotg(a+b)=\dfrac{-2\sqrt{11}+132\sqrt{2}-4\sqrt{22}}{3\cdot22}\\\sf cotg(a+b)=\dfrac{\diagdown\!\!\!2\cdot(-\sqrt{11}+66\sqrt{2}-2\sqrt{22})}{3\cdot\diagdown\!\!\!2\cdot11}\\\sf cotg(a+b)=\dfrac{-\sqrt{11}+66\sqrt{2}-2\sqrt{22}}{33}$

$\underline{\rm portanto}\\\boxed{\boxed{\boxed{\boxed{\sf cotg(arc~sec(-3)+arc~sen(0,1))=\dfrac{-\sqrt{11}+66\sqrt{2}-2\sqrt{22}}{33}}}}}\blue{\checkmark}$