If sin (α+β)=1, sin (α-β)=(1/2), then tan (α+2 β) tan (2 α+β), α, β ∈(0, π / 2), is equal to

Question

If sin (α+β)=1, sin (α-β)=(1/2), then tan (α+2 β) tan (2 α+β), α, β ∈(0, π / 2), is equal to (A) 1 (B) -1 (C) 0 (D) none of these

Tardigrade Admin · Accepted Answer

sin (α+β)=1 or α+β=(π/2) sin (α-β)=(1/2) or (α-β)=(π/6) Solving, we get α=π / 3 and β=π / 6. Now tan (α+2 β) tan (2 α+β) = tan ((π/3)+(π/3)) tan ((2 π/3)+(π/6)) = tan (2 π/3) tan (5 π/6) =(- cot (π/3))(- cot (π/6)) =(-(1/√3))(-√3) =1

Q. If sin(α+β)=1,sin(α−β)=21​, then tan(α+2β) tan(2α+β),α,β∈(0,π/2), is equal to

1

-1

0

none of these

sin(α+β)=1 or α+β=2π​sin(α−β)=21​ or (α−β)=6π​ Solving, we get α=π/3 and β=π/6. Now tan(α+2β)tan(2α+β) =tan(3π​+3π​)tan(32π​+6π​) =tan32π​tan65π​ =(−cot3π​)(−cot6π​) =(−3​1​)(−3​) =1

Q. If $sin (α + β) = 1, sin (α - β) = \frac{1}{2}$ , then $tan (α + 2 β)$ $tan (2 α + β), α, β \in (0, π /2)$ , is equal to