Find the equation for the tangent line to the curve at the point defined by the given value of t.also find d²y/dx² at this point x=SECt, y=TANt,where t=π/4.

To find the equation of the tangent line to the curve at the given point, we need to determine the slope of the curve at that point and then use the point-slope form of a line.

Let's start by finding the derivative of y with respect to x.
To do this, we need to use the chain rule since x is a function of t.

Given:
x = sec(t)
y = tan(t)
t = π/4

Using the chain rule, we have:
dy/dx = (dy/dt) / (dx/dt)

Let's find the derivative of x with respect to t:
dx/dt = d(sec(t))/dt

The derivative of sec(t) can be found using the identity d(sec(t))/dt = sec(t) * tan(t).
Since t = π/4,
dx/dt = sec(π/4) * tan(π/4) = sqrt(2) * 1 = sqrt(2)

Now, let's find the derivative of y with respect to t:
dy/dt = d(tan(t))/dt

The derivative of tan(t) can be found using the identity d(tan(t))/dt = sec^2(t).
Since t = π/4,
dy/dt = sec^2(π/4) = 2

Now, we can find the derivative of y with respect to x:
dy/dx = (dy/dt) / (dx/dt) = 2 / sqrt(2) = sqrt(2)

This is the slope of the curve at the point (x, y) = (sec(π/4), tan(π/4)).

Next, let's find d²y/dx², which is the second derivative of y with respect to x.
To do this, we need to find d(dy/dx)/dx.
We already know dy/dx = sqrt(2), so:

d(dy/dx)/dx = d(sqrt(2))/dx

Taking the derivative of sqrt(2) with respect to x gives us 0 since sqrt(2) is a constant.

Therefore, d²y/dx² = 0.

Now, let's find the equation of the tangent line using the point-slope form.

We have the slope of the tangent line, m = sqrt(2), and the point (x, y) = (sec(π/4), tan(π/4)).
Using the point-slope form:

(y - y1) = m(x - x1)

Plugging in the values, we have:
(y - tan(π/4)) = sqrt(2)(x - sec(π/4))

Simplifying the equation, we get:
y - 1 = sqrt(2)(x - sqrt(2)/2)

This is the equation of the tangent line to the curve at the given point.

To find the equation for the tangent line to the curve at the point (x, y) defined by the value of t, we need to follow these steps:

1. Step 1: Calculate dy/dx by differentiating y with respect to x.
2. Step 2: Substitute the given value of t into x and y to find the corresponding (x, y) coordinates.
3. Step 3: Substitute the value of dy/dx and the coordinates (x, y) into the point-slope formula to determine the equation of the tangent line.

Now let's go through these steps.

Step 1: Calculate dy/dx by differentiating y with respect to x.

Given that x = sec(t) and y = tan(t), we need to express y in terms of x to differentiate it.

To do this, we can use the trigonometric identity: tan(t) = sin(t) / cos(t).

Therefore, y = sin(t) / cos(t).

Next, we differentiate y with respect to x using the chain rule:

dy/dx = (dy/dt) / (dx/dt).

Now, we can differentiate y with respect to t and x with respect to t separately.

dy/dt = (d/dt)(sin(t) / cos(t))
= (cos(t) + sin(t)) / cos²(t).

dx/dt = (d/dt)(sec(t))
= sec(t)tan(t).

Plugging these values back into dy/dx:

dy/dx = [(cos(t) + sin(t)) / cos²(t)] / sec(t)tan(t)
= (cos(t) + sin(t)) / cos(t)sin(t).

Step 2: Substitute the given value of t into x and y to find the corresponding (x, y) coordinates.

Given t = π/4, we can substitute this value into x and y.

x = sec(π/4) = √2

y = tan(π/4) = 1

So, the point (x, y) is (√2, 1).

Step 3: Substitute the value of dy/dx and the coordinates (x, y) into the point-slope formula to determine the equation of the tangent line.

Using the point-slope formula, which states that the equation of a line with slope m passing through the point (x₁, y₁) is given by:

y - y₁ = m(x - x₁),

Substituting the point (√2, 1) and dy/dx = (cos(t) + sin(t)) / cos(t)sin(t) into the point-slope formula:

y - 1 = [(cos(t) + sin(t)) / cos(t)sin(t)](x - √2).

This is the equation for the tangent line to the curve at the point (x, y) = (√2, 1) for the given values of t (t = π/4).