Given:
3sin(α+β)=2sin(α−β)
⇒sin(α−β)sin(α+β)=32
⇒sin(α+β)−sin(α−β)sin(α+β)+sin(α−β)=2−32+3
⇒2cosαsinβ2sinαcosβ=−5
⇒tanβtanα=−5
⇒tanα=−5tanβ
⇒k=−5
For α,β∈(0,2π) let 3sin(α+β)=2sin(α−β) and a real number k be such that tanα=tanβ. Then the value ofk is equal to
Held on 30 Jan 2024 · Verified 6 Jul 2026.
−5
5
32
−32
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