Tan (-pi/2)=what?

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Answer 1 · 2018-03-04T13:49:54

See explanation

By the definition of tangent

$tan (x) = \frac{sin (x)}{cos (x)}$ $tan(x)=sin(x)/cos(x)$

So for $tan (- \frac{π}{2})$ $tan(-pi/2)$

$tan (- \frac{π}{2}) = \frac{sin (- \frac{π}{2})}{cos (- \frac{π}{2})} = - \frac{1}{0}$ $tan(-pi/2)=sin(-pi/2)/cos(-pi/2)=-1/0$

Which is undefined (division by 0)

However the limit must be

$lim_(x->(-pi/2)^+)=tan(x)=sin(x)/cos(x)=-oo$

$lim_(x->(-pi/2)^-)=tan(x)=sin(x)/cos(x)=oo$

Which we also can see if we graph tangent

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