Question

Se $tgx =a,\frac{\pi }{2}\ \textless \ x\ \textless \ \pi$ ,é correto afirmar que $senx + cos x$ vale;


A)$\frac{-1-a}{\sqrt{1+a^{2} } }$

B)$\frac{1+a}{\sqrt{1+a^{2} } }$

C)$\frac{1-a}{\sqrt{1+a^{2} } }$

D)$\frac{-1+a}{\sqrt{1+a^{2} } }$

E)$\frac{-1-a}{\sqrt{1-a^{2.} } }$

Answer 1

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$\sf tg(x)=a\implies tg^2(x)=a^2\\\sf sec^2(x)=tg^2x+1\\\sf sec(x)=-\sqrt{a^2+1}\\\sf cos(x)=-\dfrac{1}{\sqrt{a^2+1}}\\\sf cotg(x)=\dfrac{1}{a}\implies cotg^2(x)=\dfrac{1}{a^2}\\\sf cosec^2(x)=cotg^2(x)+1\\\sf cosec(x)=\sqrt{\dfrac{1}{a^2}+1}=\dfrac{\sqrt{a^2+1}}{a}\\\sf sen(x)=\dfrac{a}{\sqrt{a^2+1}}$

$\sf sen(x)+cos(x)=\dfrac{a}{\sqrt{a^2+1}}-\dfrac{1}{\sqrt{a^2+1}}=\dfrac{-1+a}{\sqrt{a^2+1}}\green\checkmark$