Find an equation of the tangent line to the curve at the given point. y=secx, (pi/3, 2)

Answer:Equation of tangent is [tex]y-2\sqrt{3}x=2-2\frac{\pi}{\sqrt{3}}[/tex]Step-by-step explanation:Given: Equation of curve, y = sec xPassing through point = [tex](\frac{\pi}{3},2)[/tex]We need to find Equation of tangent to the given curve and at the given point.First we find the slope of the tangent by differentiating the given curve.As we know that slope of the tangent, m = [tex]\frac{\mathrm{d}y}{\mathrm{d} x}[/tex]So, considery = sec x[tex]\frac{\mathrm{d}y}{\mathrm{d} x}=\frac{\mathrm{d}(sec\,x)}{\mathrm{d} x}[/tex][tex]\frac{\mathrm{d}y}{\mathrm{d} x}=sec\,x\:tan\,x[/tex]Now we find value of slope at given point,put x = [tex]\frac{\pi}{3}[/tex] in above derivatewe get[tex]\frac{\mathrm{d}y}{\mathrm{d} x}=sec\,\frac{\pi}{3}\:tan\,{\pi}{3}=2\sqrt{3}[/tex]Now using Slope point form, we haveEquation of tangent [tex]y-y_1=m(x-x_1)[/tex][tex]y-2=2\sqrt{3}(x-\frac{\pi}{3})[/tex][tex]y=2\sqrt{3}x-2\sqrt{3}\frac{\pi}{3}+2[/tex][tex]y-2\sqrt{3}x=2-2\frac{\pi}{\sqrt{3}}[/tex]Therefore, Equation of tangent is [tex]y-2\sqrt{3}x=2-2\frac{\pi}{\sqrt{3}}[/tex]