For α , β ∈ 0 , π 2 let 3 sin ( α + β ) = 2 sin ( α - β ) and a real number k be such that tan α = tan β .…

For α,β0,π2 let 3sin(α+β)=2sin(α-β) and a real number k be such that tanα= tanβ. Then the value of k is equal to
  1. -5
  2. 5
  3. 23
  4. -23

Solution

Given:

 3sinα+β=2sinα-β

sinα+βsinα-β=23

sinα+β+sinα-βsinα+β-sinα-β=2+32-3

2sinαcosβ2cosαsinβ=-5

tanαtanβ=-5

tanα=-5tanβ

k=-5

Asked in: JEE Main 2024 (30 Jan Shift 2)

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