## anonymous 5 years ago Find an equation of the tangent line to the curve at the given point. y = csc (x) - 2 sin (x) P=(pi/6,1)

1. anonymous

first finding the derivative of equation y'=-csc(x)cot(x) - 2cos(x) <---also the slope equation for tangent lines at every point on the graph Given the point (pi/6,1) Finding slope m=y'(pi/6)= -csc(pi/6)cot(pi/6) - 2cos(pi/6) = -3squrtroot(3) Equation for tangent line y-y1=m(x-x1) y-1=-3squrtroot(3)(x-pi/6) y= [-3\sqrt{3} + \pi \sqrt{3}/2 +1\]

2. anonymous

$y= -3\sqrt{3}x + \pi \sqrt{3}/2 +1$