Question

calcule o valor da expressão cossecx-senx sobre secx-senx , sabendo que cotgx= 5 sobre 2 e o<×< PI sobre dois

Answer 1

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:

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

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

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