What is the equation of the tangent line of #r=tan^2(theta) - sin(theta-pi)# at #theta=pi/4#?

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Answer 1 · 2018-05-10T00:01:09

#r=(2+sqrt2)/2#

#r=tan^2 theta- sin(theta - pi)# at #pi/4#

#r=tan^2 (pi/4) - sin (pi/4 -pi)#

#r=1^2 - sin((-3pi)/4)#

#r=1-sin((5pi)/4)#

#r=1-(-sqrt2/2)#

#r=1+sqrt2/2#

#r=(2+sqrt2)/2#