If cot(270° - ϕ) + tan(90° + ϕ) = 4 and 0° < ϕ < 90°, then, find the value of (tanϕ + cotϕ)?

- 2√5
- 2√3
- 3√5
- 4√5

Option 1 : 2√5

**Given:**

cot(270° - ϕ) + tan(90° + ϕ) = 4 and 0° < ϕ < 90°

**Formula Used:**

**(**a + b)^{2} = a^{2} + b^{2} + 2ab

**Calculation:**

cot(270° - ϕ) + tan(90° + ϕ) = 4

⇒ tanϕ – cotϕ = 4

Squaring both sides –

⇒ tan2ϕ + cot2ϕ – 2tanϕ cotϕ = 16

⇒ tan2ϕ + cot2ϕ – 2 = 16

⇒ tan2ϕ + cot2ϕ = 18

⇒ tan2ϕ + cot2ϕ + 2tanϕ cotϕ = 18+ 2tanϕcotϕ

⇒ (tanϕ + cotϕ)2 = 20

⇒ tanϕ + cotϕ = √20

⇒ tanϕ + cotϕ = 2√5

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