#f(theta)= 5 theta sin (3 theta)+2 cot (11 theta); theta= pi/12#

#f(pi/12)= 5*pi/12* sin(3 *pi/12)+2 *cot (11*pi/12); pi/12 ~~ 0.26#

#f(pi/12)~~ - -6.54 # The point at which tangent to be drawn is

# (x_1=0.26 ,y_1= -6.54) # Slope of the curve is

#f'(theta)= 5 sin (3 theta)+15 theta cos (3 theta)-22 csc^2 (11 theta); #

Slope of the curve at #theta= pi/12# is

#f'(pi/12)= 5 sin (3 * pi/12)+15 * pi/12 * cos(3 * pi/12)-22 * csc^2 (11* pi/12); #

#f'(pi/12)= m ~~ -322.11# . Equation of tangent at

# (0.26 ,-6.54)#

is #(y-y_1=m(x-x_1) or y + 6.54 = -322 .11 (x-0.26)# or

# y= -322.11 x +84.33 -6.54 or y= -322.11 x +77.79#

Equation of tangent is #y= -322.11 x +77.79# [Ans]