Question

# Which of the following statements are correct? 1 . if sin θ = sin α ⇒ θ = nπ + (−1)nα where α ∈ [ - π2 , π2] 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 ∈ (- π2 , π2 ) n ∈​ I

A

Only 1 & 2

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B

Only 1 & 3

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C

Only 2 & 3

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D

All 1, 2 & 3

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Solution

## The correct option is D All 1, 2 & 3 1.) sinθ = sin α sinθ = sin α = 0 2. cos (θ+α2) .sin (θ−α2) = 0 cos (θ+α2) or sin (θ−α2) = 0 (θ+α2) = (2n+1)π2 or (θ−α2) = 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)2nα ,where n ∈ Z Combining these two results we get, x = nπ + (−1)nα, where n ∈ Z 2.) cos θ = cos α cos θ - cos α = 0 2 sinθ+α2.sinθ−α2 = 0 -2 sinθ+α2.sinθ−α2 = 0 sinθ+α2 = 0 or .sinθ−α2 = 0 θ+α2 = nπ or θ−α2 = nπ θ = 2nπ - α or θ = 2nπ + α Hence θ = 2nπ ± α, where n ∈ Z 3 )tanθ = tan α sinθcosθ - sinαcosα = 0 sinθ.cosα−cosθ.sinαcosθ.cosα = 0 sin(θ−α)cosθ.cosα = 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

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