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  • Usuário do Brainly
2014-08-22T00:28:25-03:00
Observe que, \text{sec}^2~x=1+\text{tg}^2~x

Temos que, \text{tg}~x-\text{sec}~x=-21.

Assim, \text{tg}~x=\text{sec}~x-21.

Elevando os dois lados ao quadrado, obtemos \text{tg}^2~x=\text{sec}^2~x-42\text{sec}~x+441.

Por outro lado, \text{tg}^2~x=\text{sec}^2-1.

Logo:

\text{sec}^2~x-1=\text{sec}^2~x-42\text{sec}~x+441

42\text{sec}~x=442

\text{sec}~x=\dfrac{442}{42}

\text{sec}~x=\dfrac{221}{21}

Assim, \text{tg}~x=\sqrt{\left(\dfrac{221}{21}\right)^2-1}=\sqrt{\dfrac{48~400}{441}}=\dfrac{220}{21}.

Portanto, \text{tg}~x+\text{sec}~x=\dfrac{220}{21}+\dfrac{221}{21}=\dfrac{220+221}{21}=21.
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