Question

What is the derivation of formula tanx=tany x=nπ+y?

Answer 1

Step-by-step explanation:

We have,

tan θ = tan ∝

⇒ sin θ/cos θ - sin ∝/cos ∝ = 0

⇒ (sin θ cos ∝ - cos θ sin ∝)/cos θ cos ∝ = 0

⇒ sin (θ - ∝)/cos θ cos ∝ = 0

⇒ sin (θ - ∝) = 0

⇒ sin (θ - ∝) = 0

⇒ (θ - ∝) = nπ, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….), [Since we know that the θ = nπ, n ∈ Z is the general solution of the given equation sin θ = 0]

⇒ θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Hence, the general solution of tan θ = tan ∝ is θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Note: The equation cot θ = cot ∝ is equivalent to tan θ = tan ∝ (since, cot θ = 1/tan θ and cot ∝ = 1/tan ∝). Thus, cot θ = cot ∝ and tan θ = tan ∝ have the same general solution.

Hence, the general solution of cot θ = cot ∝ is θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

1. Solve the trigonometric equation tan θ = 1√3

Solution:

tan θ = 1√3

⇒ tan θ = tan π6

⇒ θ = nπ + π6, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….), [Since, we know that the general solution of tan θ = tan ∝ is θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)]

2. What is the general solution of the trigonometric equation tan x + tan 2x + tan x tan 2x = 1?

Solution:

tan x + tan 2x + tan x tan 2x = 1

tan x + tan 2x = 1 - tan x tan 2x

tanx+tan2x1−tanxtan2x = 1

tan 3x = 1

tan 3x = tan π4

3x = nπ + π4, where n = 0, ± 1, ± 2, ± 3,…….

x = nπ3 + π12, where n = 0, ± 1, ± 2, ± 3,…….

Therefore, the general solution of the trigonometric equation tan x + tan 2x + tan x tan 2x = 1 is x = nπ3 + π12, where n = 0, ± 1, ± 2, ± 3,…….