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GIVEN:
2 cosθ = √3
(2cosθ + 1)/(2cosθ - 1) = a + b√3
SOLUTION:
2 cosθ = √3 [Given]
==> cosθ = √3/2
==> θ = 30⁰
(2cosθ + 1)/(2cosθ - 1) = a + b√3 [Given]
∵ cosθ = √3/2,
==> [2(√3/2) + 1]/[2(√3/2) - 1] = a + b√3
LHS:-
==> (√3 + 1)/(√3 - 1)
==> (√3 + 1)(√3+1)/(√3 - 1)(√3 + 1)
[By Conjugate Multiplication]
==> (√3 + 1)²/(3 - 1)
==> (3 + 2√3 + 1)/2
==> (4 + 2√3)/2 [Take 2 as common factor]
==> 2(2 + √3)/2
==> 2 + √3
Now, compare this LHS to the RHS, you can see both of them look similar,
==> 2 + √3 = a + b√3
∴ a = 2, b = 1
Answer:
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