I need help on finding the local linear approximation of tan 62 degree.

i got 1.77859292096 can someone check if i got it right?

tan(60+2) = (tan 60+ tan 2) / (1- tan60 tan 2)

But tan2 appx = sin 2deg = sin2PI/180= 2PI/180

tan 62=(tan 60+2PI/180) / (1- tan60 2PI/180)

check that. It does not equal your answer.

The actual value of tan 62 is 1.880726465.
I don't know what you mean by the "local linear" approximation. If you were doing a linear Taylor series approximation around 60 degrees, you would get
tan 62 = tan 60 + [d/dx(tan x)@ 60 deg]* 2 deg*(pi/180)(deg/radian)= sqrt 3 + (8 pi/180 = 1.8716
The pi/180 factor converts degrees to radians for use in the linear approximation.

To find the local linear approximation of tan(62 degrees), you can use a Taylor series approximation centered around a nearby angle, such as 60 degrees. The local linear approximation assumes that the function can be approximated by a straight line in the vicinity of the chosen angle.

The Taylor series expansion of the tangent function is given by:

tan(x) = tan(a) + (x - a)*(d/dx(tan x)|a) + higher-order terms,

where a is the angle around which the approximation is centered and d/dx(tan x)|a is the derivative of the tangent function evaluated at a.

In this case, we are approximating tan(62 degrees) around a = 60 degrees:

tan(62 degrees) = tan(60 degrees) + (62 - 60)*(d/dx(tan x)|60).

The derivative of tan(x) is sec²(x), so we evaluate it at 60 degrees:

d/dx(tan x)|60 = sec²(60 degrees) = sec²(π/3 radians) = 4/√3.

Substituting these values into the formula, we get:

tan(62 degrees) ≈ tan(60 degrees) + (62 - 60)*(4/√3) = √3 + (4/√3) ≈ 1.732 + 2.309 ≈ 4.041.

Therefore, the local linear approximation of tan(62 degrees) is approximately 4.041.

It seems that your calculation of 1.77859292096 is not correct. The actual value of tan(62 degrees) is approximately 1.880726465.