Respostas

  • Usuário do Brainly
2014-08-18T18:47:01-03:00
Temos que:

\rhd \text{cotg}~x=\dfrac{1}{\text{tg}~x}

\rhd \text{sec}^2~x=1+\text{tg}^2~x

\rhd \text{sec}~x=\dfrac{1}{\text{cos}~x}

\rhd \text{sen}^2~x+\text{cos}^2~x=1

\rhd \text{cossec}~x=\dfrac{1}{\text{sen}~x}.

Seja A=\dfrac{\text{cossec}~x-\text{sen}~x}{\text{sec}~x-\text{sen}~x}.

Como \text{cotg}~x=\dfrac{5}{2}, podemos afirmar que, \text{tg}~x=\dfrac{2}{5}.

Assim, \text{sec}^2~x=1+\left(\dfrac{2}{5}\right)^2=\dfrac{29}{25}, donde, \text{sec}~x=\dfrac{\sqrt{29}}{5}.

Consequentemente, \text{cos}~x=\dfrac{5}{\sqrt{29}}=\dfrac{5\sqrt{29}}{29}.

Deste modo, \text{sen}^2~x+\left(\dfrac{5\sqrt{29}}{29}\right)^2=1 e obtemos:

\text{sen}~x=\sqrt{1-\dfrac{25}{29}}=\dfrac{4}{29}=\dfrac{2}{\sqrt{29}}.

Ou ainda, \text{sen}~x=\dfrac{2\sqrt{29}}{29}.

Assim, \text{cossec}~x=\dfrac{29}{2\sqrt{29}}=\dfrac{29\sqrt{29}}{58}=\dfrac{\sqrt{29}}{2}.

Logo:

A=\dfrac{\dfrac{\sqrt{29}}{2}-\dfrac{2\sqrt{29}}{29}}{\dfrac{\sqrt{29}}{5}-\dfrac{2\sqrt{29}}{29}}=\dfrac{\dfrac{29\sqrt{29}-4\sqrt{29}}{58}}{\dfrac{29\sqrt{29}-10\sqrt{29}}{145}}

A=\dfrac{\dfrac{25\sqrt{29}}{58}}{\dfrac{19\sqrt{29}}{145}}=\dfrac{25\sqrt{29}}{58}\cdot\dfrac{145}{19\sqrt{29}}

A=\dfrac{25\times145}{58\times19}=\dfrac{125}{38}