If sin(α+β)=1,sin(α−β)=1/2;α,β∈[π/2] then tan(α+2β)tan(2α+β) is equal to
We have,
sin(α+β)=1 and, sin(α−β)=12⇒ α+β=π2 and, α−β=π6⇒ α=π3,β=π6∴ tan(α+2β)tan(2α+β)=tan2π3tan5π6=−cotπ6−cotπ3=1