The equations of L_1 and L_2 are...?

Question

The equations of L_1 and L_2 are y = mx and y = nx, respectively. Suppose that L_1 makes twice as large an angle with the horizontal (measured counterclockwise from the positive x-axis) as does L_2, and that L_1 has 4 times the slope of L_2. If L_1 is not horizontal, then mn is
(A) sqrt[2]/2
(B) -sqrt[2]/2
(C) 2
(D) -2
(E) not uniquely determined by the given information

Answer 1

Let x be the angle L1 makes with x -axis
Then 2x = angle L2 makes with x-axis.
m= tan2x
n = tanx
m=4n so tan2x = 4 tanx
2tanx/(1-tan^2x) = 4tan x
2tan x = 4tanx -4tan^3x
-4tan^3x = - 2tanx
2tan^2x =1
tan^2x = 1/2
tan x = sqrt(1/2) = sqrt(2)/2 = n
tan 2x = (2*sqrt(2)/2)/(1 - (sqrt(2)/2)^2) = 2sqrt(2)=m
So mn = sqrt(2)/2 * 2sqrt(2) =2

Answer 2

given
m=4n
Ang_1 = 2Ang_2

Slope is the tangent of the line. To find the angle take the arctan of the slope.

So

arctan(m) = 2 * arctan(n)

Substitute 4n for m

arctan(4n) = 2 * arctan(n)

Now take the tan of each side of the equation

tan(arctan(4n)) = tan(2 * arctan(n))

We know tan(arctan(x) = x and tan(2 arctan(y)) = 2y/(1-y^2).

4n = 2n / (1 -n^2)
1 - n^2 = 1/2
n^2 = 1/2
n = sqrt(2)/2

m = 4n = 2sqrt(2)

mn = 2sqrt(2)*sqrt(2)/2 = 2