General Solution of Trigonometric Function (Tan θ)
General Solution of Equation tan θ = tan α:
Given that,
tan θ = tan α
\(\frac{\sin \theta }{\cos \theta }=\frac{\sin \alpha }{\cos \alpha }\),
sin θ. Cos α – sin α cos θ = 0
sin (θ – α) = 0
(θ – α) = sin⁻¹ (0)
(θ – α) = sin⁻¹ (sin nπ)
(θ – α) = nπ
θ = nπ + α, where n ϵ z
Note: For general solution of the equation tan θ = k, where k ϵ R. we have tan θ = tan⁻¹ (tan k). Thus θ = nπ + (tan⁻¹ k), n ϵ Z.
Examples 1: Solve tan 3θ = -1
Solution: Given that,
tan 3θ = -1
3θ = tan⁻¹ (-1)
3θ = tan⁻¹ (tan -π/4)
3θ = nπ – π/4, n ϵ Z
\(\theta =\frac{n\pi }{3}-\frac{\pi }{12}\), n ϵ Z
Example 2: Solve 2 tan θ – cot θ = -1
Solution: Given that,
2tan θ – cot θ = -1
\(2\tan \theta -\frac{1}{\tan \theta }=-1\),
2tan² θ + tan θ – 1 = 0
(tan θ + 1) (2tan θ – 1) = 0
tan θ = -1
tan θ = tan – π/4
θ = nπ – π/4; n ϵ Z and
2 tan θ – 1 = 0
tan θ = ½
θ = tan⁻¹ (½)
θ = nπ + tan⁻¹ (½); n ϵ Z.