The equations of the lines L 1 and L 2 are y = m x and y = n x , respectively. Suppose L 1 makes twice as large an angle with the horizontal (measured counterclockwise from the positive x -axis) as does L 2 and m = 4 n , then the value of m 2 + 4 n 2 m 2 - 6 n 2 is equal to (where, n ≠ 0 )

Step-by-step solution

Let, m = tan ⁡ 2 θ and n = tan ⁡ θ And m = 4 n and m ≠ 0 , ∴ tan ⁡ 2 θ ≠ 0 ;   θ ≠ 0 tan ⁡ 2 θ = 4 tan ⁡ θ 2 tan ⁡ θ 1 - tan 2 ⁡ θ = 4 tan ⁡ θ ∴ 1 = 2 - 2 tan 2 ⁡ θ ⇒ 2 tan 2 ⁡ θ = 1 ⇒ tan 2 ⁡ θ = 1 2 = n 2 ⇒ m 2 = 16 n 2 = 8 ⇒ m 2 + 4 n 2 m 2 - 6 n 2 = 10 5 = 2