**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.