# Find an equation of the tangent line to the curve at the given point.

## Homework Statement

Find an equation of the tangent line to the curve at the given point.

y = tan x at point (pi/4,1)

## The Attempt at a Solution

step 1. find the derivative of tan x, which sec^2 x
step 2. find the slope.
this is where I mess up. I assume that I have to square x where sec is pi/4. Sec^2 of pi/4 I think would be the hypotenuse/x which is 1/2√2. So I square that, which to me would be 1/ (4*2) or 1/8. But the solution manual says it's the square of 2√2. why? Obviously I don't know how to square secant.

step 3. insert the slope into the y - y = m(x - x) formula, which I can do.

In the solution manual below, they don't have negative signs, so there's a few negative signs missing but I know where they are. Related Calculus and Beyond Homework Help News on Phys.org
Ok, I got a new idea. The slop at pi/4,1, the coordinates are 2√2, 2√2. So maybe I should just find the slope of that, don't worry about squaring it. Well, the slop of that coordinate is 2 root 2, divided by 2 root 2, which is 1, not 2 like the book says. still lost.

(5 minutes later)

No, that can't be right, the book clearly says that you find the slope by sec^2 pi/4 = the square root of two squared. still don't see why.

Last edited:
jbunniii
Homework Helper
Gold Member
You miscalculated sec(pi/4).

sec(x) = 1/cos(x).

What is cos(pi/4)?

at pi/4, the coordinates are (2√2,2√2), so the cosine would be x/hyp, which is 2√2/1 and the secant would be

1/(2√2/1), if you multiply the numerator by the inverse of the denominator you get:

1 * (1/2√2) and if you square that you get 1/2.

when i put in my calculator cos^-1 3.14/4 i get .67 radians.

jbunniii
Homework Helper
Gold Member
at pi/4, the coordinates are (2√2,2√2), so the cosine would be x/hyp, which is 2√2/1
You should be able to see immediately that this is wrong. The cosine (and sine) of any angle must be between -1 and 1.

You have the coordinates wrong.

now I got it. i still have troubles getting the concepts of trig right. i was confusing the coordinates with the angle.

cos (45) = .52, sec (45) = 1.92 = square that and you almost get 4, so I guess it's right.

jbunniii
Homework Helper
Gold Member
I'm not sure what you mean by almost 4. That's not the answer I got for the slope. Your method in the previous post looked correct, you just got the coordinates wrong.

At pi/4, the unit circle coordinates are (1/sqrt(2), 1/sqrt(2)).

So what's sec(pi/4)?

jbunniii