Find the equation of the line tangent to curve x=sec(t), y=tan(t), at t=pi/6.

Question

I found th slope(derivative of the two which was csc(t). Plugged in Pi/6 and got 2 as the slope.

Now how do I find the equation of the tangent? There are like two equations to work with, I'm confused.

Answer 1

at t=pi/6, y=1/√3 and x=2/√3

So, now you have a point (2/√3,1/√3) and a slope (2), so the line is (using the point-slope form)

y - 1/√3 = 2(x - 2/√3)