What is the equation of the straight line tangent to the graph of y = tan x at the point (4 ; 1)?

Our task is to find the equation of the straight line tangent to
the graph of  y = tan x at the point (4 ; 1)

I would use trigo
identity to convert tangent into sine/consine:

y=tan x= (sin x ) /
(cos x)

To find the equation of the tangent to any curve, we find
the 1st derivative of the
curve:

dy/dx

= [ (cos x) . d (sin x)/
dx  - (sin x) . d (cos x)/dx ] / (cos x)^2

= [ (cos x). (cos x)  -  
(sin x) . (- sin x) ] / (cos x)^2

= [ (cos x)^2  +  (sin x)^2 ] /
(cos x)^2

= 1 / (cos x)^2

To find the
slope of the tangent to the curve at (4, 1), we substitute  x=4  into the expression 1 / (cos
x)^2 .

1 / (cos x)^2

=  1 / (cos
4)^2

=  2.3406

We can find the value of
c by putting this value into "m" of the standard equation of a straight line y = mx + c ,
together with the values of x=4 and y=1.

y = mx +
c

1 = (2.3406)(4) + c

c =
1-(2.3406)(4)

=  1-9.3624

=  
-8.3624

Hence the equation of the tangent to the curve at (4,1)
is

y = 2.3406x -
8.3624

To be convinced of the solution, click on the
link provided at the bottom of this
answer