We have some long solution here. I will try to shorten it.
From the given #r=4 cos((3pi)/2-theta)+sin (-theta)#
this equation is reducible to
#r=4[cos ((3pi)/2) cos(theta)+ sin((3pi)/2) sin theta]-sin theta#
also because #sin (-theta)=-sin theta#
#r=4[cos ((3pi)/2) cos(theta)+ sin((3pi)/2) sin theta]-sin theta#
#r=4 [0*cos theta+(-1) sin theta]-sin theta#
#r=-4 sin theta-sin theta#
#r=-5 sin theta#
Solve for the point of tangency: given #theta=(-5pi)/12#
#r=-5 sin theta=-5 sin ((-5pi)/12)=5/4(sqrt6+sqrt2)#
The polar coordinate of the point is
#(r, theta)=(5/4(sqrt6+sqrt2), (-5pi)/12)#
We will use rectangular coordinates for convenience
#x=r*cos theta# and #y=r*sin theta#
Solve x
#x=r*cos theta#
#x=5/4(sqrt6+sqrt2)*cos ((-5pi)/12)#
#x=(5(sqrt6+sqrt2))/4((sqrt6-sqrt2))/4#
#x=5/4#
Solve y
#y=r*sin theta#
#x=5/4(sqrt6+sqrt2)*sin ((-5pi)/12)#
#x=5/4(sqrt6+sqrt2)(-(sqrt6+sqrt2)/4)#
#x=-5/16(8+4sqrt3)#
#x=-5/4(2+sqrt3)#
We now have the rectangular coordinate point
#(x, y)=(5/4, -5/4(2+sqrt3))#
We need the slope #m# on the curve #r=-5 sin theta#
For convenience , we convert this to rectangular coordinate system
#r=-5 sin theta#
#sqrt(x^2+y^2)=-5 *y/sqrt(x^2+y^2)#
Simplify
#(sqrt(x^2+y^2))^2=-5 *y/cancelsqrt(x^2+y^2)*cancelsqrt(x^2+y^2)#
#x^2+y^2=-5y#
#x^2+y^2+5y=0#
Solve for m by differentiating implicitly
#x^2+y^2+5y=0#
#2x+2yy'+5y'=0#
#y'(2y+5)=-2x#
#y'=(-2x)/(2y+5)#
#m=(-2x)/(2y+5)=(-2(5/4))/(2( -5/4(2+sqrt3))+5)#
#m=(-(5/2))/(( -5/2(2+sqrt3))+5)#
#m=1/sqrt3#
Finally, let us determine the tangent line: using Two-Point Form
#y-y_1=m(x-x_1)#
#y--5/4(2+sqrt3)=1/sqrt3(x-5/4)#
#color(red)(y+5/4(2+sqrt3)=1/sqrt3(x-5/4)" ")#The Tangent Line
kindly see the graph of #r=4[cos ((3pi)/2) cos(theta)+ sin((3pi)/2) sin theta]-sin theta# and #y+5/4(2+sqrt3)=1/sqrt3(x-5/4)#the Tangent Line
graph{(y+5/4(2+sqrt3)-1/sqrt3(x-5/4))(x^2+y^2+5y)=0[-20,20,-10,10]}
God bless....I hope the explanation is useful.
