Question

Which of the following statements are correct?

1 . if sin θ = sin α ⇒ θ = nπ + $(-1{)}^n$α where α ∈ [ - $\frac{\pi }{2}$ , $\frac{\pi }{2}$] n ∈ I

2 . if cos θ = cos α ⇒ 2nπ±α where α ∈ [0,π] n ∈​ I

3 . if tan θ = tan α ⇒ θ = nπ + α where α ∈ (- $\frac{\pi }{2}$ , $\frac{\pi }{2}$ ) n ∈​ I

• A. Only 1 & 2
• B. Only 1 & 3
• C. Only 2 & 3
• D. All 1, 2 & 3

Answer 1

The correct option is D

All 1, 2 & 3

1.) sinθ = sin α

sinθ = sin α = 0

2. cos ($\frac{\theta +\alpha }{2}$) .sin ($\frac{\theta -\alpha }{2}$) = 0

cos ($\frac{\theta +\alpha }{2}$) or sin ($\frac{\theta -\alpha }{2}$) = 0

($\frac{\theta +\alpha }{2}$) = (2n+1)$\frac{\pi }{2}$ or ($\frac{\theta -\alpha }{2}$) = nπ where n ∈ Z

θ = (2n+1) π - α or θ = 2nπ - α where n ∈ Z

Hence

θ = (2n+1) π + $(-1{)}^{2n+1}$ × α or θ = 2nπ + $(-1{)}^{2n}$α ,where n ∈ Z

Combining these two results we get,

x = nπ + $(-1{)}^n$α, where n ∈ Z

2.) cos θ = cos α

cos θ - cos α = 0

2 sin$\frac{\theta +\alpha }{2}$.sin$\frac{\theta -\alpha }{2}$ = 0

-2 sin$\frac{\theta +\alpha }{2}$.sin$\frac{\theta -\alpha }{2}$ = 0

sin$\frac{\theta +\alpha }{2}$ = 0 or .sin$\frac{\theta -\alpha }{2}$ = 0

$\frac{\theta +\alpha }{2}$ = nπ or $\frac{\theta -\alpha }{2}$ = nπ

θ = 2nπ - α or θ = 2nπ + α

Hence θ = 2nπ ± α, where n ∈ Z

3 )tanθ = tan α

$\frac{sin\theta }{cos\theta }$ - $\frac{sin\alpha }{cos\alpha }$ = 0

$\frac{sin\theta .cos\alpha -cos\theta .sin\alpha }{cos\theta .cos\alpha }$ = 0

$\frac{sin(\theta -\alpha )}{cos\theta .cos\alpha }$ = 0

sin(θ-α) = 0 and θ & α should NOT be odd multiples if π/2

θ-α = nπ

θ = nπ + α where n ∈ Z

θ & α ≠ (2n+1) π/2