If tanθ + cotθ = 4/√3, where 0 < θ < π/2, then sinθ + cosθ is equal to

(a) 1
(b) (√3 – 1)/2
(c) (√3 + 1)/2
(d) √2

Anurag Mishra Professor Asked on 20th February 2016 in Maths.
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  • 1 Answer(s)

    Answer: (c) (1 + √3)/2

    Solution:-
    tanθ + cotθ = 4/√3
    tanθ + 1/tanθ  = 4/√3
    = (tan²θ + 1)/tanθ
    As we know that tan²θ + 1 = sec²θ,

    Then, sec²θ/tanθ = 4/√3
    cosθ/cos²θ sinθ = 4/√3
    1/cosθ sinθ  = 4/√3
    Multiply and divide by 2,
    Then, 2/2cosθ sinθ = 4/√3                              (sin2θ = 2 sinθcosθ)
    1/sin2θ = 2/√3
    Then, sin2θ = √3/2
    sin2θ = sin60º
    2θ = 60
    θ = 60/2 => θ = 30º

    So, the value of Sinθ + cosθ = sin30º + cos30º
    1/2 + √3/2 = (1 + √3)/2

    Hence, the correct answer is option (c) (1 + √3)/2. 

    Anurag Mishra Professor Answered on 20th February 2016.
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