you need to find the slope of f(x) at #x=pi/4# and the point #(pi/4,f(pi/4))#
1) finding slope:
#f'(x) = d/dx(sinx) - d/dx(x/cosx)#
#f'(x) = cosx - (cosxd/dx(x) - xd/dx(cosx))/((cosx)^2)# (trig differentiation rule and quotient rule)
#f'(x) = cosx - (cosx(1) - x(-sinx))/((cosx)^2)#
#f'(x) = cosx - (cosx + xsinx)/((cosx)^2)#
plug in #x=pi/4#:
#f'(pi/4) = cos(pi/4) - (cos(pi/4)+pi/4sin(pi/4))/((cos(pi/4))^2)#
#f'(pi/4) = 1/sqrt(2) - (1/sqrt(2)+pi/4*1/sqrt(2))/((1/sqrt(2))^2)#
#f'(pi/4) = 1/sqrt(2) - ((4+pi)/(4sqrt(2)))/(1/2)#
#f'(pi/4) = 2/(2sqrt(2)) - (4+pi)/(2sqrt(2)#
#f'(pi/4) = (-2-pi)/(2sqrt(2))# = tangent line slope
2) finding point:
#f(pi/4) = sin(pi/4)-(pi/4)/cos(pi/4)#
#f(pi/4) = sqrt(2)/2 - (pi/4)/(1/sqrt(2))#
#f(pi/4) = (2sqrt(2))/4 - pisqrt(2)/4#
#f(pi/4) = (2sqrt(2)-pisqrt(2))/4#
3) the tangent line is: #y - f(pi/4) = slope*(x - pi/4)#
#y - (2sqrt(2)-pisqrt(2))/4 = (-2-pi)/(2sqrt(2))(x-pi/4)#
you can convert to slope-intercept if you want
