Question

# Let be such that and Then cannot satisfy

Hint:

### 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:

### Here we have to find the which region ϕ cannot satisfy.

Now we have,

Given equation

⇒2cosθ (1−sin ϕ) =sin^{2}θ(x cos ϕ)−1 [since, tan x + cot y = 2/sin2x]

⇒2cosθ−2cossin 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 = .

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