How do you find an equation of the tangent line to the curve at the given point #y=secx - 2cosx# and #(pi/3, 1)#?
1 Answer
Apr 25, 2018
Explanation:
#"the slope of the tangent m "=dy/dx" at "x=pi/3#
#rArrdy/dx=secxtanx+2sinx#
#dy/dx(x=pi/3)=sec(pi/3)tan(pi/3)+2sin(pi/3)#
#color(white)(xxxxxxxxx)=2xxsqrt3+2xxsqrt3/2#
#color(white)(xxxxxxxxx)=2sqrt3+sqrt3=3sqrt3#
#rArry-1=3sqrt3(x-pi/3)#
#rArry=3sqrt3x-pisqrt3+1larrcolor(red)"equation of tangent"#
