Answer :

α = tan¯¹ (7 + 4√3)

Solution :

tan α = |(√3+1)|/|(√3-1)|

tan α = (√3+1)²/(√3-1)²

tan α = (3 + 1 + 2√3) ÷ (3 + 1 - 2√3)

tan α = (4 + 2√3) ÷ (4 - 2√3)

Rationalising the denominator

(4 + 2√3)(4 + 2√3) ÷ (4 - 2√3)(4 + 2√3)

(4 + 2√3)² ÷ (4² - (2√3)²)

(16 + 12 + 16√3) ÷ (16 - 12)

(28 + 16√3) ÷ 4

7 + 4√3

tan α = 7 + 4√3

α = tan¯¹ (7 + 4√3)

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