What is the slope of the tangent line of $r=(2theta+sin2theta)/cos^2theta$ at $theta=(-3pi)/8$ ?

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Answer 1 · 2017-04-16T22:59:10

$"m"_"tan"=16+8sqrt2+(12+9sqrt2)pi$

$(dr)/(d theta)=((cos^2theta)(2+2cos2theta)+(2costhetasintheta)(2theta+sin2theta))/cos^4(theta)$

$r'(-3/8pi)=$
$((cos^2(-3/8pi))(2+2cos2(-3/8pi))+(2cos(-3/8pi)sin(-3/8pi))(2(-3/8pi)+sin2(-3/8pi)))/cos^4((-3/8pi))$

$=16+8sqrt2+(12+9sqrt2)pi$

What is the slope of the tangent line of r=(2theta+sin2theta)/cos^2thetar=(2theta+sin2theta)/cos^2theta at theta=(-3pi)/8theta=(-3pi)/8?