Question

Tangent Line h(t) = sec t / t @ (Pi, 1/pi)

Answer 1

Take derivitive

note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of $\frac{f(x)}{g(x)} = \frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}$
so

the derivitive of $\frac{sec(t)}{t} = \frac{sec(t)tan(t)t-(1)(sec(t))}{t^2}$
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get
$\frac{(-1)(0)(pi)-(pi)(-1)}{pi^2}= \frac{pi}{pi^2}= \frac{1}{pi}$
slope=1/pi

use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)

y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1