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

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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

y= tanx ( 4,1)

Let the equation of the line be :

y- y1 = m (x-x1) where (x1,y1) is any point on the line and m is the slope.

==> ( y- 1) = m (x - 4)

Now we will calculate the slope:

We know that the slope of the tangent line equals the derivative at the point .

==> y' = sec^2 x

==> sec^2 4 = 1.004 = 1 ( approx)

Then the equation of the line is:

y- 1 = 1( x-4)

y-1 = x - 4

y= x-4 + 1

**y= x-3**

To find the tangent to y= tanx at (4,1).

Tangent to y = f(x) at x1, y1 is given by:

y-y1 = {f'(x) at (x1,y1)} (x-x1).

Here y = tanx. x1 = 4. So y1 = tan4.

dy/dx = sec^2x. So {dy/dx at (4, 1)} = sec^2 1.

Therefore the equation of tangent at (4,1) is given by:

y -tan4 = (sec 1)^2 {x-tan1)

Therefore y - tan4 = (sec 1)^2 {x-tan1} is the equation of the tangent at (4,1).