Respuesta :

LammettHash
[tex]\sec t=\dfrac1{\cos t}[/tex]

and

[tex]\cos(n\pi)=(-1)^n[/tex]

for any integer [tex]n[/tex]. With [tex]n=-3[/tex], we get [tex]\cos(-3\pi)=(-1)^{-3}=-1[/tex], so

[tex]\sec(-3\pi)=\dfrac1{-1}=-1[/tex]

\sec t=\dfrac1{\cos t}

and

\cos(n\pi)=(-1)^n

for any integer n. With n=-3, we get \cos(-3\pi)=(-1)^{-3}=-1, so

\sec(-3\pi)=\dfrac1{-1}=-1