CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

Only 1 & 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

Only 2 & 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

All 1, 2 & 3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D

All 1, 2 & 3


1.) sinθ = sin α

sinθ = sin α = 0

2. cos (θ+α2) .sin (θα2) = 0

cos (θ+α2) or sin (θα2) = 0

(θ+α2) = (2n+1)π2 or (θα2) = 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.sinθα2 = 0

-2 sinθ+α2.sinθα2 = 0

sinθ+α2 = 0 or .sinθα2 = 0

θ+α2 = nπ or θα2 = nπ

θ = 2nπ - α or θ = 2nπ + α

Hence θ = 2nπ ± α, where n ∈ Z

3 )tanθ = tan α

sinθcosθ - sinαcosα = 0

sinθ.cosαcosθ.sinαcosθ.cosα = 0

sin(θα)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


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General Solutions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon