Question

How do you find an equation of the tangent line to the curve at the given point `y=secx - 2cosx` and `(pi/3, 1)`?

Answer 1

`y=3sqrt3x-pisqrt3+1`

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"`