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\[\huge f'(x)=1+cosx\]

f'(x) = m

now plug in x=pi into the x-value

yeah.. now put in x=pi.. that would give you the slope.

y= 180

\[\large f'(\pi )= 1 + \cos(\pi) = 0\]

m=0

because cos(pi) = -1.

you won, madame. :)

y=180 would be the equation ?

y=180

-_- y = pi.
now we have m =0, y = pi, x= pi.
\[y-y_{1}= m(x-x_{1})\]\[y-\pi = 0(x-\pi)\]\[y=\pi\]

No .. we can substitute pi for 180 @Jhannybean

You get the same thing :P