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Easy

Question

Let theta comma ϕ element of left square bracket 0 comma 2 pi right square bracket be such that 2 cos space theta left parenthesis 1 minus sin space ϕ right parenthesis equals sin squared space theta open parentheses tan space theta over 2 plus co t space space theta over 2 close parentheses cos space ϕ minus 1 comma tan space left parenthesis 2 pi minus theta right parenthesis greater than 0 and negative 1 less than sin space theta less than negative fraction numerator square root of 3 over denominator 2 end fraction Then ϕ cannot satisfy

  1. 0 less than ϕ less than pi over 2
  2. pi over 2 less than ϕ less than fraction numerator 4 pi over denominator 3 end fraction
  3. fraction numerator 4 pi over denominator 3 end fraction less than ϕ less than fraction numerator 3 pi over denominator 2 end fraction
  4. fraction numerator 3 pi over denominator 2 end fraction less than ϕ less than 2 pi

hintHint:

In this question, we have to find the which region ϕ cannot satisfy. Now firstly, solve the given equation and find the value θ and ϕ and find the range of ϕ . If the given option is cannot match with the range then it will be our answer.

The correct answer is: 0 less than ϕ less than pi over 2


    Here we have to find the which region ϕ cannot satisfy.
    Now we have,
    Given equation 2 cos space theta left parenthesis 1 minus sin space ϕ right parenthesis equals sin squared space theta open parentheses tan space theta over 2 plus co t space space theta over 2 close parentheses cos space ϕ minus 1
    ⇒2cosθ (1−sin ϕ) =sin2θ(fraction numerator 2 over denominator sin theta end fractionx cos ϕ)−1 [since, tan x + cot y = 2/sin2x]
    ⇒2cosθ−2costhetasin ϕ =2sinθcos ϕ −1
    ⇒2cosθ+1=2sin(θ+ ϕ) .....(i)
    Also given that tan(2π−θ)>0
    ⇒tanθ<0.....(1)
    −1<sinθ<− √3/2
    ⇒θ ϵ (3Π/2, 5Π/3) .....(2)
    So, ′θ′ is in 4th quadrant ⇒ L.H.S. of equation (i) will be positive.
    1<2cosθ+1<2
    ⇒1<2sin(θ+ ϕ)<2
    ⇒21<sin(θ ϕ)<1
    ⇒2π+π/6 < θ + ϕ < 5π/6 +2π
    ⇒ 2π +π/6 − θmax < ϕ <2π+ 5π/6 −θmin
    ⇒ π/2 < ϕ < 4π/3
    Therefore, ϕ satisfy itself from π/2 to 4π/3 .
    The correct answer is 0 < ϕ < 4π/3 , 3π/2 < ϕ < 2π and 4π/3 < ϕ < 3π/2.

    In this question we have to find the the which region ϕ cannot satisfy. In this question more than one option is correct. Here firstly solve the given equation. Remember that, , tan x + cot x    = fraction numerator 2 over denominator sin x end fraction.

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